ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 4, Issue 1 (2014), Pages 231-262
Editors: Valmir Krasniqi, Jos´e Luis D´ıaz-Barrero, Armend Sh. Shabani, Paolo Perfetti, Mohammed Aassila, Mih´aly Bencze, Valmir Bucaj, Emanuele Callegari, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba, Cristinel Mortici, Jozsef S´andor, Ercole Suppa, David R. Stone, Roberto Tauraso, Francisco Javier Garc´ıa Capit´an.
PROBLEMS AND SOLUTIONS
Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction. Proposals should be accompanied by solutions. An asterisk (*) indicates that nei-ther the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a com-mittee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]
Solutions to the problems stated in this issue should arrive before June 15, 2014
Problems
88. Proposed by Hun Min Park, Korea Advanced Institute of Science and Technol-ogy, Daejeon, South Korea. Suppose that three real numbers a, b, c(0 ≤ a, b, c, ≤ 1) satisfy the following equality;
X cyc a − b 1 − ab· a 1 − a2 = 0 Prove that a = b = c. (Note thatX cyc
means ’cyclic sum’X cyc
f (x, y, z) = f (x, y, z) + f (y, z, x) + f (z, x, y))
89. Proposed by Mohammed Aassila, Strasbourg, France.
Let S be the set of positive integers that does not contain the digit 7 in their decimal representation. Prove that
X n∈S
1
n < +∞.
c
2010 Mathproblems, Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e. 231
90. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technol-ogy, Damascus, Syria. Let n be a positive integer, prove that
n X k=0 (−1)b √ kc ≤√n , and determine the cases of equality.
91. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. Calculate Z 1 0 Z 1 0 ln(1 − x) ln(1 − xy)dxdy.
92. Proposed by D.M. B˘atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzˇau, Romania. Let {an}n≥0be a sequence of positive integer numbers such that 5 does not divide anfor all, n ∈ N,and let the sequence {bn}n≥0be defined by bn= a2nL2n, for n ∈ N where {Ln}n≥0is the sequence of Lucas numbers. Prove that bnis square-free, (i.e. bn is not a perfect square), for every n ∈ N \ {1}.
93. Proposed by Anastasios Kotronis, Athens, Greece. For x ∈ (−1, 1), evaluate
+∞ X n=1 (−1)n+1n tan−1x − x +x 3 3 − · · · + (−1) n+1x2n+1 2n + 1 .
94. Proposed by Sava Grozdev and Deko Dekov (Jointly), Bulgaria. Find a point P in the plane of a given triangle ABC, such that the sum
|AP |2 b2 + |BP |2 c2 + |CP |2 a2 is minimal, where a = BC, b = CA and c = AB.
95. Proposed by Li Yin, Department of Mathematics, Binzhou University, Binzhou City, Shandong Province, 256603, China. An approximation formula of Wallis product. For all n ∈ N, then
Wn∼ 1 √ π 1 −ln n 2n n
Solutions
No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.
81. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technol-ogy, Damascus, Syria. Find, in terms of a > 0, the minimum of
a(x2+ y2+ z2) + 9xyz xy + yz + zx
when x, y and z are nonnegative real numbers such that x + y + z = 1..
Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Answer: The minimum is 2a if a ≤ 1 and a + 1 if a ≥ 1.
Proof Let x + y + z = 3u, xy + yz + zx = 3v2, xyz = w3. Let m be the searched minimum. We homogenize by writing
a(x2+ y2+ z2)(x + y + z) + 9xyz (xy + yz + zx)(x + y + z) which in terms of the variables (u, v, w) becomes
a(9u2− 6v2)3u + 9w3
9uv2 − m ≥ 0
that is
9w3+ R(u, v) ≥ 0
This is a linear function in w3 and then it holds if and only if it holds for the minimum value of w3. The standard theory says that, once fixed the valued of (u, v), the minimum value of w3 occurs when at least one among the variables is zero or when two of the are equals.
In the first case we set z = 0 and get a(x2+ y2)
xy ≥ m
and it is evident that m = 2a.
As for the second case we set x = y and get a(x2+ y2+ z2)(x + y + z) + 9xyz
(xy + yz + zx)(x + y + z) − a − 1 =
(x − z)2(2ax − 2x + az) x(x + 2z)(2x + z)
which is positive for any x, z ∈ R if and only if a ≥ 1. The minimum is thus the maximum between 2a and a + 1 which is 2a if a ≤ 1 while a + 1 if a ≥ 1.
Solution 2 by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia. a) Let a ≥ 1. Schur’s inequality is equivalent to
x2+ y2+ z2+ 9xyz
Using the above inequality and the given condition x + y + z = 1, we have
x2+ y2+ z2+ 9xyz ≥ 2(xy + yz + zx) (1) Also clearly, we have x2+ y2+ z2≥ xy + yz + zx. From this we find that
(a − 1)(x2+ y2+ z2) ≥ (a − 1)(xy + yz + zx). (2) Add (1) and (2) we get
a(x2+ y2+ z2) + 9xyz ≥ (a + 1)(xy + yz + zx) or equivalently
a(x2+ y2+ z2) + 9xyz
xy + yz + zx ≥ a + 1. This inequality becomes equality when x = y = z = 13. Hence
min a(x
2+ y2+ z2) + 9xyz xy + yz + zx
= a + 1.
b) Let 0 < a < 1. From (1) we have,
a(x2+ y2+ z2) + 9axyz ≥ 2a(xy + yz + zx). (3) On the other hand we have
a(x2+ y2+ z2) + 9xyz ≥ a(x2+ y2+ z2) + 9axyz (4) From (3) and (4) we have
a(x2+ y2+ z2) + 9xyz ≥ 2a(xy + yz + zx) or equivalently
a(x2+ y2+ z2) + 9xyz xy + yz + zx ≥ 2a. with equality when x = y = 12, and z = 0. Hence
min a(x
2+ y2+ z2) + 9xyz xy + yz + zx
= 2a.
Also solved by Arkady Alt, San Jose, California, USA; Moti Levy, Re-hovot, Israel; and the proposer.
82.Proposed by Anastasios Kotronis, Athens, Greece. Let F (n) :=P
k≥1 1
(kn+1)k!. Determine the sequence {cm}m≥0 such that lim n→+∞n mF (n) − m−1 X k=0 ck nk = cm
where, for m = 0, the sum is considered to be 0.
Solution 1 by Moti Levy, Rehovot, Israel. Actually, the problem deals with finding asymptotic expansion of F (n) =R1
0 e xn− 1 dx. 1 kn + 1 = 1 kn 1 1 +kn1 = ∞ X j=1 (−1)j−1 kj 1 nj
∞ X k=1 1 (kn + 1)k! = ∞ X k=1 ∞ X j=1 (−1)j−1 kjk! 1 nj = ∞ X j=1 1 nj ∞ X k=1 (−1)j−1 kjk! . Let {cm}m≥0= ( (−1)m−1 ∞ X k=1 (−1)m−1 kmk! ) m≥0 then ∞ X k=1 1 (kn + 1)k! = c1 n + c2 n2 + c3 n3 + · · · .
The sequence {cm} can be expressed by the generalized hypergeometric function, cm= (−1)
m−1
m+1Fm+1(1, 1, . . . , 1; 2, 2, . . . , 2; 1) . For m = 1, in particular,
c1=2F2(1, 1; 2, 2; 1) = −γ + Ei (1) ∼= 1.3179
Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy.
F (n)=. X k≥1 1 (kn + 1)k! = X k≥1 1 k!kn ∞ X j=0 (−1)j (kn)j
The absolute convergence of the series allows us to take the limit n → +∞ under the sum so for m = 0 we have
lim n→+∞F (n) = 0, (c0= 0) Let m = 1. nm F (n) − m−1 X k=0 ck nk ! = nF (n) =X k≥1 1 k!k ∞ X j=0 (−1)j (kn)j
whose limit n → +∞ is evidently c1 = ∞ X k=1
1
k!k. By induction we suppose that
cm= ∞ X k=1
(−1)m−1
k!km for any 0 ≤ m ≤ r. We have
nr+1 F (n) − r X m=0 cm nm ! = nr+1 ∞ X j=0 (−1)j nj+1 X k≥1 1 k!kj+1 − r X m=1 cm nm = nr+1 r X p=1 (−1)p−1 np X k≥1 1 k!kp + ∞ X p=r+1 (−1)p−1 np X k≥1 1 k!kp − r X m=1 cm nm
and the limit as n → ∞ yields (−1)r ∞ X k=1
1 k!kr+1.
Solution 3 by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. Let Ω = {z ∈ C : <z > −1}. For z ∈ Ω we consider G(z) defined by the formula
G(z) = ∞ X p=1 1 p! (p + z)
Clearly, this is a series of analytic functions on Ω (namely: z 7→ 1/(k! (z + k)),) that converges to G uniformly on every compact subset of Ω. This proves that G, itself, is analytic in Ω and that, for every m ≥ 0 and every z ∈ Ω, we have
G(m)(z) = ∞ X p=1 (−1)mm! p! (p + z)m+1
In particular, for |z| < 1 we have
G(z) = ∞ X m=0 G(m)(0) m! z m
Thus for z in the neighborhood of 0, and for every m ≥ 0 (with the same convention as in the statement of the problem,) we have
zG(z) = m−1
X k=0
ckzk+ O(zm)
with c0= 0 and ck = G(k−1)(0)/(k − 1)! when k is a positive integer. But F (n) = 1
nG( 1
n), so for large n, and for every nonnegative integer m, we have
F (n) = m−1 X k=0 ck nk + O 1 nm .
Moreover, according to (1) we have
ck = (−1)k−1 ∞ X p=1 1 p! pk.
This yields the desired conclusion. Note that the ck’s do not seem to have a closed form. but they can, alternatively, be expressed as integrals:
ck = (−1)k−1 (k − 1)! ∞ X p=1 1 p! Z ∞ 0 tk−1e−ptdt = (−1) k−1 (k − 1)! Z ∞ 0 tk−1ee−t − 1dt.
Also solved by Moubinool Omarjee, Paris, France, and the proposer.
Editor’s Comment: The proposer of this problem indicated that it is a general-ization of problem U278 of Mathematical Reflections.
83.Proposed by Hun Min Park, Korea Advanced Institute of Science and Technol-ogy,Daejeon, South Korea. Let {an} strictly increasing sequence of positive integers such that gcd(ai, aj) = 1 for any i, j(i < j). Let bn = an+1− an. Prove that the sequence {bn} is unbounded(has no upper bound).
Solution 1 by Shmuel Isaac and Moti Levy, Rehovot, Israel (Jointly). Let us define the sequence {pn}, where pn is the largest prime number that divides an. Since ai and aj are relatively prime (for i 6= j) then pi 6= pj. Let A (x) be the number of all {an} sequence terms, which are less or equal to x, i.e.,
aA(x)≤ x, and aA(x)+1> x.
Let P (x) be the number of {pn} sequence terms, which are less or equal to x. By the definitions above,
A (x) ≤ P (x) ≤ π (x) , (1)
where π (x) is the number of prime numbers less or equal to x.
Let X be a positive integer, X ≥ 2a1. By definition of A (x) ,we have aA(X)+1> X, hence aA(X)+1− a1≥ X − X 2 = X 2 . Clearly, A(X) X n=1 bn= aA(X)+1− a1.
Suppose, on the contrary, that the sequence {bn} is bounded by the positive con-stant M > 0. Then aA(X)+1− a1≤ A (X) M, X 2 ≤ aA(X)+1− a1≤ A (X) M. (2) By equations ((1)) and (2), π (X) X ≥ A (X) X ≥ 1 2M Multiplying both sides by ln X > 0,
π (X) X/ ln X ≥
1 2M ln X.
On one hand, the Prime Number Theorem states that limX→∞ π(X)
X/ ln X = 1 but on the other hand limX→∞ 2M1 ln X = ∞; This is a contradiction, which leads us to the conclusion that the sequence {bn} is unbounded.
Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let pnbe the smallest prime number that divides an. Since gcd(ai, aj) = 1 for every distinct i and j we conclude that p1, p2, . . . , pn are distinct primes from the interval [1, an] and consequently, π(an) ≥ n. where π(x) represents the number of primes p that are smaller or equal to x. Now, according to the weak form of the prime number theorem, there exists an absolute positive constant A such that π(x) ≤ A x
ln x hence n ≤ π(an) ≤ A an ln an ≤ A an ln n
where we used the trivial inequality n ≤ an (since, with a0 = 0, we have an = Pn
k=1(ak− ak−1) ≥P n
k=11 = n.) Thus, an≥n ln nA , and in particular lim
n→∞ an
Now, suppose that the sequence {bn} is bounded by some constant M then it follows immediately that
an− a1 n = 1 n n−1 X k=1 bk≤ M
and consequently lim supn→∞ann ≤ M which is a absurd according to (3). This contradiction proves that {bn} cannot be bounded.
Remark. Refining upon the proof presented above we see that we have
lim inf n→∞
an
n ln n ≥ 1, and lim supn→∞ bn ln n > 1.
Solution 3 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. The sequence {aj} with the smallest entries aj for any j is the sequence of the prime numbers. Any other sequence {a0j} is such that a0j≥ aj and then asymptotically a0j ≥ c0j ln j for any j large enough. We argue by contradiction by assuming that there exists a constant c1 such that bn ≤ c1. This implies that
n X k=1
aj+1− aj= an+1− a1≤ nc1 but this contradicts a0j ≥ c0j ln j if n is large enough. Also solved by Arthur Handle, and the proposer.
84.Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, China.
Let 1 < p < ∞, we can generalize the inverse of arcsin as follows:
arcsinp(x) = Z x 0 1 (1 − tp)1/pdt, 0 ≤ x ≤ 1 and πp 2 = arcsinp(1) = Z 1 0 1 (1 − tp)1/pdt. The inverse of arcsinp(x) on 0,
πp 2
is called the generalized sine function and denoted by sinp. The generalized cosine function is defined as cosp(x) = dxd sinp(x). Similarly, the generalized inverse hyperbolic sine function
arcsinhp(x) = Z x 0 1 (1 + tp)1/pdt , x ∈ [0, ∞), −arcsinhp(−x) , x ∈ (−∞, 0)
generalizes the classical inverse hyperbolic sine function. The inverse of arcsinhp is called the generalized hyperbolic sine function and denoted by sinhp. The gener-alized hyperbolic cosine function is defined as coshp(x) = sinh0p(x). For p > 2 and x ∈ (0,πp 2 ), prove that ln x sinpx <sinpx − x cospx p sinpx and lnsinhpx x > x coshpx − sinhpx p sinhpx .
85.Proposed by D.M. B˘atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzˇau, Romania. Let n be a positive integer. Prove that
Xn k=1 Fk2 Lk · n X k=1 Fk3 L2 k ≥(Fn+2− 1) 5 (Ln+2− 3)2
where Fn , respectively Ln represents the nth Fibonacci number respectively the nthLucas number.
Solution 1 by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia. Using the following well known indentities L1+ L2+ · · · + Ln = Ln+2− 3 and F1+ F2+ · · · + Fn= Fn+2− 1 we have n X k=1 Lk !3 · n X k=1 F2 k Lk ! n X k=1 F3 k L2 k ! ≥ n X k=1 Fk !5 . (1)
It is enough to prove the above inequality. Applying the H¨older’s inequality, we get
n X k=1 Lk !3/5 · n X k=1 Fk2 Lk !1/5 · n X k=1 Fk3 L2 k !1/5 = n X k=1 L3/5k 5 3 !3/5 · n X k=1 F2 k Lk 1/5!5 1/5 · n X k=1 F3 k L2 k 1/5!5 1/5 ≥ n X k=1 L3/5k · F 2/5 k L1/5k · Fk3/5 L2/5k ! = n X k=1 Fk. Hence (1) is proved.
Solution 2 by Moti Levy, Rehovot, Israel. A more general version of this problem appeared in ”The Fibonacci Quarterly”, Volume 51, Number 4, November 2013. n X k=1 Fkm+1 Lm k ! n X k=1 Fkp+1 Lpk ! ≥ (Fn+2− 1) m+p+2 (Ln+2− 3) m+p , m > 0, p > 0. (2)
To prove (2), we use the inequality (3) due to J. Radon,
n X k=1 xp+1k ykp ≥ (Pn k=1xk) p+1 (Pn k=1yk) p , p > 0, and xk ≥ 0, yk> 0, for 1 ≤ k ≤ n. (3) By Radon’s inequality, n X k=1 Fkm+1 Lm k ! n X k=1 Fkp+1 Lpk ! ≥ ( Pn k=1Fk) p+m+2 (Pn k=1Lk) p+m (4)
The following sums are well known, n X k=1 Fk= Fn+2− 1 (5) n X k=1 Lk= Ln+2− 3 (6)
The required result is obtained by substituting (5) and (6) in (3).
Solution 3 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. We will use the following lemma:
Lemma. Let p, λ, µ be real numbers with p > 1, and 0 < λ, µ < 1. Then for every positive numbers a1, . . . , an and b1, . . . , bn we have:
(a1+ · · · + an)2p (b1+ · · · + bn)2p−2 ≤ n X k=1 a2λpk b2µ(p−1)k · n X k=1 a2(1−λ)pk b2(1−µ)(p−1)k .
Proof. Let x1, . . . , xn and y1, . . . , yn be positive numbers such that n X k=1 xk= n X k=1 yk = 1
Let r > 1 be defined by 1p+1r = 1, then by H¨older’s inequality we have
1 = n X k=1 xk= n X k=1 xk yk1/r · yk1/r≤ n X k=1 xpk yp/rk !1p n X k=1 yk !r1 = n X k=1 xpk ykp−1 !1p
That is, by the Cauchy-Schwarz inequality:
1 ≤ n X k=1 xpk ykp−1 = n X k=1 xλpk yµ(p−1)k · x (1−λ)p k y(1−µ)(p−1)k ≤ v u u t n X k=1 x2λpk y2µ(p−1)k · n X k=1 x2(1−λ)pk yk2(1−µ)(p−1)
for 0 < λ, µ < 1. Applying this, with xk = a/˜a and yk/˜b with ˜ a = a1+ · · · + an and ˜b = b1+ · · · + bn, we obtain (a1+ · · · + an)2p (b1+ · · · + bn)2p−2 ≤ n X k=1 a2λpk b2µ(p−1)k · n X k=1 a2(1−λ)pk b2(1−µ)(p−1)k . Taking, (p, λ, µ) = 52,35,23, we obtain (a1+ · · · + an)5 (b1+ · · · + bn)3 ≤ n X k=1 a3k b2 k ! · n X k=1 a2k bk ! .
Finally, the desired inequality is obtained, by setting ak= Fk, bk= Lk and noting that n X k=1 Fk = n X k=1 (Fk+2− Fk+1) = Fn+2− F2= Fn+2− 1, n X k=1 Lk = n X k=1 (Lk+2− Lk+1) = Ln+2− L2= Ln+2− 3.
Remark. Similarly, for m > 2 and 0 < q, p < m, we have n X k=1 Fkm−p Lm−qk ! · n X k=1 Fkp Lq−2k ! ≥ (Fn+2− 1) m (Ln+2− 3)m−2 .
with the same proof.
Also solved by G. C. Greubel, Newport News, VA, USA; ´Angel Plaza, University of De Las Palmas, Grain Canaria, Spain; and the proposer.
86.Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj, Romania. Calculate Z 1 0 ln(√x +√1 − x) √ x dx.
Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Let x = t2. The integral reads as
Z 1 0 2 ln(t +p1 − t2)dt = 2 Z 1 0 ln tdt + 2 Z 1 0 ln 1 + r 1 t2 − 1 ! dt Z 1 0 ln t dt = (t ln t − t) 1 0= −1
In the second integral we change t = 1/p1 + y2 and it becomes Z ∞ 0 ln(1 + y) y (1 + y2)3/2dy = − ln(1 + y) p 1 + y2 ∞ 0 + Z ∞ 0 1 1 + y 1 p 1 + y2dy = |{z} y=sinh t Z ∞ 0 dt 1 + sinh t|{z}= z=et 2 Z ∞ 1 1 z2+ 2z − 1dz Z ∞ 1 1 z − z1 − 1 z − z2 1 z1− z2 dz where z1= (−1 + √ 2)/2, z2= (−1 − √ 2)/2. We get evidently −1 √ 2ln 1 − z1 1 − z2 = −1√ 2ln 2 −√2 2 +√2 = −1 √ 2ln(3 − 2 √ 2) = ln(3 + 2 √ 2) √ 2 and the integral finally is
√
2 ln(3 + 2√2) − 2.
Solution 2 by Arkady Alt, San Jose, California, USA.
Let I :=R01ln √
x +√1 − x √
x dx. The change of variables x = sin 2
t shows that
I = 2 Z π/2
0
ln (sin t + cos t) cos t dt.
= 2 Z π/2
0
ln (sin t + cos t) sin t dt. (t ← π 2 − t)
Taking the half sum we obtain
I = Z π/2
0
ln (sin t + cos t) (cos t + sin t)dt.
=√2 Z π/4 −π/4 ln√2 cos θcos θ dθ (t ← π4 + θ) = √ 2 2 (ln 2) Z π/4 −π/4 cos θ dθ | {z } I1 +√2 Z π/4 −π/4 ln(cos θ) cos θ dθ | {z } I2 Clearly, I1= √ 2, and I2= sin θ ln(cos θ) iπ/4 −π/4+ Z π/4 −π/4 sin2θ cos θ dθ =√2 ln√1 2 + 2 Z 1/ √ 2 0 u2 1 − u2du (u = sin θ) = − √ 2 2 ln 2 − √ 2 + Z 1/ √ 2 0 1 1 + u+ 1 1 − u du = − √ 2 2 ln 2 − √ 2 + ln √ 2 + 1 √ 2 − 1 = − √ 2 2 ln 2 − √ 2 + 2 ln(√2 + 1) Finally I = −2 + 2√2 ln(√2 + 1).
Also solved by Albert Stadler, Switzerland; Omran Kouba, Higher In-stitute for Applied Sciences and Technology, Damascus, Syria; Anas-tasios Kotronis, Athens, Greece; G. C. Greubel, Newport News, VA, USA; Moti Levy, Rehovot, Israel; Moubinool Omarjee, Paris, France, AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; and the proposer.
87.Proposed by Dorlir Ahmeti, University of Prishtina, Republic of Kosova. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that
√ a +√b 1 +√ab + √ b +√c 1 +√bc + √ c +√a 1 +√ca ≥ 3.
Solution 1 by AN-anduud Problem Solving Group. The proposed problem is equivalent to the following problem. x, y, z be positive real number such that x2+ y2+ z2= 3. Prove that x + y 1 + xy+ y + z 1 + yz+ z + x 1 + zx ≥ 3. (1)
Applying H¨older’s inequality, we have x + y 1 + xy + y + z 1 + yz + z + x 1 + zx 2
((x + y)(1 + xy)2+ (y + z)(1 + yz)2+ (z + x)(1 + zx)2)
Thus (1) would follow from (2) If we prove the next inequality
8(x + y + z)3≥ 9((x + y)(1 + xy)2+ (y + z)(1 + yz)2+ (z + x)(1 + zx)2) or equivalntly
9(x5+ y5+ z5) + 48xyz ≥ 25(x3+ y3+ z3). (3) To prove (3) we note that
9X cyc x5+ 48xyz − 25X cyc x3=X cyc (x + y)(x − y)2(x + y − z)2 +X cyc z(x − y)2((x + y − 6z)2+ 35xy) + 25xyzX cyc (x − y)2≥ 0
Also solved by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; and the proposer.
MATHCONTEST SECTION
This section of the Journal offers readers an opportunity to solve interesting and el-egant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals are always wel-comed. The source of the proposals will appear when the solutions be published.
Proposals
61. Find all real solutions of the following system of equations: p
x2+ y2+ 6x + 9 +px2+ y2− 8y + 16 = 5, 9y2− 4x2= 60.
62. Let P be an interior point to an equilateral triangle ABC. Draw perpendiculars P X, P Y and P Z to the sides BC, CA and AB, respectively. Compute the value of
BX + CY + AZ P X + P Y + P Z
63. How many ways are there to weigh of 31 grams with a balance if we have 7 weighs of one gram, 5 of two grams, and 6 of five grams, respectively?
64. Let A(x) be a polynomial with integer coefficients such that for 1 ≤ k ≤ n + 1, holds:
A(k) = 5k Find the value of A(n + 2).
65. Let a0, a1, . . . , an and b0, b1, · · · , bn be complex numbers. If n ≥ 2, then prove that Re n X k=0 akbk ! ≤ 1 2n n + 1 n n X k=0 |ak|2+ n2 n − 1 2n − 2 n n X k=0 |bk|2 !
Solutions
56. Find all positive integers n smaller that 201314 such that 3n ≡ 3 (mod 13) and 5n≡ 5 (mod 13). What are the smallest and the biggest? How many are there in total?
(50th Catalonian Mathematical Olympiad)
Solution by Eloi Torrent Juste, AULA Escola Europea, Barcelona, Spain. We have that 30 ≡ 1 (mod 13), 31 ≡ 3 (mod 13), 32 ≡ 9 (mod 13), 33 ≡ 1 (mod 13), and so forth. Thus, powers of three when divided by 13 present a cycle of order 3. Likewise, 50≡ 1 (mod 13), 51≡ 5 (mod 13), 52≡ 12 (mod 13), 53≡ 8 (mod 13), 54≡ 1 (mod 13), etc. So, powers of five when divided by 13 present a cycle of order 4. The integers n searched are of the form n = 3k + 1 and at the same time of the form n = 4h + 1. Therefore, n − 1 must be multiple of 3 and 4 at the same time. That is, a multiple of 12 and n = 12k + 1. According to the statement 1 ≤ 12k + 1 ≤ 201314 and therefore 0 ≤ k ≤ 201313
12 or 0 ≤ k ≤ 16776. The smallest positive integer, for k = 0 is 1. The biggest, for k = 16776 is 201313 and the total number is 16776 + 1 = 16777.
2 Also solved by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain.
57. Let n, m be positive integers. Prove that 1 + 1 n n < 1 + 1 m m+1
(Training Sessions of Catalonian Team for OME 2013)
Solution by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Applying AM-GM inequality, namely
n √
a1a2. . . an<
a1+ a2+ . . . + an
n ,
where a1, a2, . . . , an are positive numbers not all the same; we have
n+m+1 s 1 + 1 n n 1 − 1 m + 1 m+1 < n 1 + 1 n + (m + 1) 1 − 1 m + 1 n + m + 1 = 1
From the preceding, we have
1 + 1 n n 1 − 1 m + 1 m+1 < 1 ⇔ 1 + 1 n n 1 + 1 m −(m+1) < 1
Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let f be the function defined for x > −1 by f (x) = ln(1 + x), and let a be a positive real number. Using the Mean Value Theorem, there is a real number ξ ∈ (0, a) such that
ln(1 + a) a = f (a) − f (0) a = f 0(ξ) = 1 1 + ξ ∈ 1 1 + a, 1 .
that is, for a > 0, we have a
1 + a < ln(1 + a) < a.
Applying the upper inequality with a = 1/n and the lower one with a = 1/m, we get n ln 1 + 1 n < 1 < (m + 1) ln 1 + 1 m
Taking exponentials yields the desired inequality.
Also solved by Arkady Alt, San Jose, California, USA, Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy, and Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain.
58. Let x = 2 cos A, y = 2 cos B and z = 2 cos C, where A, B, C are the measures of the angles of an acute triangle ABC. Find the minimum value of
x4+ y4+ z4+ x2y2z2
(Training Catalonian Team for OME 2014)
Solution 1 by Arkady Alt, San Jose, California, USA. By replacing (cos A, cos B, cos C) in identity
cos2A + cos2B + cos2C + 2 cos A cos B cos C = 1
withx 2, y 2, z 2 we obtain x2 4 + y2 4 + z2 4 + 2 · x 2 · y 2 · z 2 = 1 Thus x2+ y2+ z2+ xyz = 4
Since triangle ABC is acute, that is x, y, z > 0, then by QM-AM inequality we have x4+ y4+ z4+ x2y2z2 4 ≥ x2+ y2+ z2+ xyz 4 2 = 1 =⇒ x4+ y4+ z4+ x2y2z2≥ 4.
Since the lower bound 4 can be attained if x = y = z ⇐⇒ A = B = C, the desired minimum is 4.
Solution 2 by Jos´e Luis D´ıaz-Barrero BARCELONA TECH, Barcelona, Spain. Since A + B + C = π, then we have
x2+ y2+ z2+ xyz = 4 cos2A + 4 cos2B + 4 cos2(A + B) − 8 cos A cos B cos(A + B)
= 4(cos2A + cos2A − cos2A cos2A + sin2A sin2B)
= 4hsin2B(cos2A + sin2A) + cos2Bi= 4
Taking into account AM-QM inequality yields
1 = x 2+ y2+ z2+ xyz 4 ≤ r x4+ y4+ z4+ x2y2z2 4
from which follows x4 + y4+ z4+ x2y2z2 ≥ 4. So, the minimum value of the expression claimed is 4 and it is attained when 4ABC is equilateral.
Solution 3 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. The answer is 4 and it is attained only when the triangle is equilateral.
First, note as in the preceding solutions we have x2+ y2+ z2+ xyz = 4 Thus,
x4+ y4+ z4+ x2y2z2− 4 = x4+ y4+ z4+ x2y2z2− 2(x2+ y2+ z2+ xyz) + 4 = (x2− 1)2+ (y2− 1)2+ (z2− 1)2+ (xyz − 1)2≥ 0 with equality if and only if x2= y2= z2= xyz = 1, or equivalently A = B = C = 60◦.
Remark. Note that the condition that ABC is acute is unnecessary.
Solution 4 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. It is a known standard result that 1 ≤ cos A + cos B + cos C ≤ 3/2 and then 2 ≤ 2 cos A + 2 cos B + 2 cos C ≤ 3. The minimum 1 cor-responds to a degenerate isosceles triangle while the maximum to an equilateral triangle. Clearly we have 2 ≤ x + y + z ≤ 3. We employ the so called ”uvw” theory which can be found at The art of problem solving forum. Define three new quantities
x + y + z = 3u, xy + yz + zx = 3v2, xyz = w3 We have
x4+y4+z4+x2y2z2= (w3)2+12uw3+81u4−108u2v2+18v4= (w3)2+12uw3+R(u, v) This is a convex increasing parabola if w3 ≥ 0 whose minimum has negative ab-scissa. It follows that the minimum of the parabola occurs when w = 0 or when w is minimum once fixed the values of u and v. According to the theory, the latter occurs when x = y (or cyclic). If w = 0 we have for instance z = 0 that is C = π/2. At x + y fixed, the minimum of x4+ y4, occurs when x = y that is A = B = π/4 or z = y =√2. This yields
If z = y and x = a − 2y, a ≤ 2 ≤ 3 we get x4+ y4+ z4+ (xyz)2− 3a 3 4 −a 3 6 =(3y − a) 2 729 P (y, a) =(3y − a) 2 729 (324y
4− 108ay3+ 1458y2− 1620ay + 702a2− 27a2y2− 6a3y − a4)
Now we prove that P (y, a) > 0. Indeed
702a2− a4= a2(702 − a2) ≥ a2(702 − 9) = 693a2, 6a3y = 6a2ay ≤ 54ay 1011y2+ 693a2> 1674ay = (1620 + 54)ay
so we get
324y4+ 442y2≥ 108ay3+ 27a2y2 and this is implied by
324y4+ 442y2≥ 108 · 3y3+ 27 · 9y2 ⇐⇒ 324y4+ 199y2≥ 324y3 and this finally follows by the AGM 324y4+ 199y2≥ 507y3. We have showed that
x4+ y4+ z4+ (xyz)2− 3a 3 4 −a 3 6 ≥ 0
and the difference equals zero if x = y = z = a/3. On account of their definition, x, y, z can be equal if and only if A = B = C = π/6 whence x = y = z = 1 and
x4+ y4+ z4+ (xyz)2− 4 ≥ 0 The searched minimum is thus min{8, 4} = 4.
Also solved by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain.
59. Let x, y, z be nonzero complex numbers. Prove that
81 1 |x|2 + 1 |y|2 + 1 |z|2 −1 ≤ 3|x + y + z|2+ |2x − y − z|2+ |2y − x − z|2+ |2z − x − y|2
(Training Catalonian Team for OME 2014)
Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. More generally, consider n nonzero complex numbers z1, . . . , zn, and let µ = 1n(z1+ . . . + zn). Clearly we have
n X k=1 |zk− µ|2= n X k=1 |zk|2− 2 n X k=1 <(zkµ) + n X k=1 |µ|2 = n X k=1 |zk|2− 2n|µ|2+ n|µ|2= n X k=1 |zk|2− n|µ|2 That is n X k=1 |zk|2= n|µ|2+ n X k=1 |zk− µ|2 (1)
Now, the HM -AM inequality shows that n2 n X k=1 1 |zk|2 !−1 ≤ n X k=1 |zk|2 (2)
Combining (1), (2), and rearranging, we obtain
n4 n X k=1 1 |zk|2 !−1 ≤ n n X k=1 zk 2 + n X k=1 (n − 1)zk− X j6=k zj 2 .
The proposed inequality corresponds to n = 3 and (z1, z2, z3) = (x, y, z).
Solution 2 by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. We have x + y + z 3 2 +1 3 2x − y − z 3 2 + 2y − x − z 3 2 + 2z − x − y 3 2! = (x + y + z)(x + y + z) 9 +1 3 (2x − y − z)(2x − y − z) 9 + (2y − x − z)(2y − x − z) 9 + (2z − x − y)(2z − x − y) 9 =1 3 |x| 2+ |y|2+ |z|2
Applying AM-HM inequality, yields 1 3 |x| 2+ |y|2+ |z|2 ≥ 1 3 |x|2+ 1 |y|2 + 1 |z|2 and taking into account that the given inequality is equivalent to
3 1 |x|2 + 1 |y|2 + 1 |z|2 ≤1 9(|x + y + z| 2) +1 27 |2x − y − z|2+ |2y − x − z|2+ |2z − x − y|2
from which the statement follows. Notice that equality holds when |x| = |y| = |z| and we are done.
Solution 3 by Arkady Alt, San Jose, California, USA. Since |x + y + z|2= (x + y + z) (x + y + z)
= |x|2+ |y|2+ |z|2+ (xy + xy) + (yz + yz) + (zx + zx) , |2x − y − z|2= (2x − y − z) (2x − y − z)
= 4 |x|2+ |y|2+ |z|2− 2 (xy + xy) + (yz + yz) − (zx + zx) , |2y − x − z|2= (2y − z − x) (2y − z − x)
= |x|2+ 4 |y|2+ |z|2− 2 (yz + yz) + (zx + zx) − (xy + xy) , |2z − x − y|2= (2z − x − y) (2z − x − y)
then
3 |x + y + z|2+ |2x − y − z|2+ |2y − z − x|2+ |2z − x − y|2
= 3|x|2+ |y|2+ |z|2+ 6|x|2+ |y|2+ |z|2= 9|x|2+ |y|2+ |z|2 and the original inequality becomes
81 1 |x|2+ 1 |y|2 + 1 |z|2 !−1 ≤ 9|x|2+ |y|2+ |z|2 or 9 1 |x|2 + 1 |y|2 + 1 |z|2 !−1 ≤|x|2+ |y|2+ |z|2, where latter inequality holds because by Cauchy’s inequality
|x|2+ |y|2+ |z|2 1 |x|2 + 1 |y|2 + 1 |z|2 ! ≥ 9
Also solved by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain.
60. Compute the following sum:
X 1≤i1≤n+1 1 i1 + X 1≤i1<i2≤n+1 1 i1i2 + . . . + X 1≤i1<...<in≤n+1 1 i1i2. . . in
(Training Spanish Team for VJIMC 2014) Solution 1 by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain. Let us denote by S the following sum:
S = 1 + X 1≤i1≤n+1 1 i1 + X 1≤i1<i2≤n+1 1 i1i2 + . . . + 1 1 · 2 · · · n + 1 = n+1 Y k=1 1 + 1 k = n+1 Y k=1 k + 1 k = (n + 2)! (n + 1)! = n + 2 Then X 1≤i1≤n+1 1 i1 + X 1≤i1<i2≤n+1 1 i1i2 + . . . + X 1≤i1<...<in≤n+1 1 i1i2. . . in = S − 1 + 1 1 · 2 · · · n + 1 = n + 1 − 1 (n + 1)!
and we are done. 2
Solution 2 by Arkady Alt, San Jose, California, USA. Consider the poly-nomial P (x) = n+1 Y k=1 x + 1 k = xn+1+ an+1+ n X k=1 akxn+1−k
Then by general Vieta’s Theorem
ak =
X 1≤i1<i2<...<ik≤n+1
1 i1i2. . . ik
, 1 ≤ k ≤ n + 1
Since an+1= 1
(n + 1)! then original sum is a1+ a2+ . . . + an= P (1) − 1 − 1 (n + 1)!. From the other hand
P (1) = n+1 Y k=1 1 + 1 k = n+1 Y k=1 k + 1 k = n + 2 Hence, a1+ a2+ . . . + an= (n + 2) − 1 − 1 (n + 1)! = n + 1 − 1 (n + 1)!.
Solution 3 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let Sn be the proposed sum and let Nn+1 = {1, 2, . . . , n + 1}. We will use the following formula:
n+1 Y k=1 (1 + bk) = n+1 X k=0 X X⊂Nn+1 |X|=k Y j∈X bj ,
that can be proved by induction. Taking bj = 1/j and noting that terms cor-responding to X = ∅ and X = Nn+1 are missing in the proposed sum, we see that Sn= n+1 Y k=1 1 + 1 k − 1 − 1 (n + 1)! = n + 1 − 1 (n + 1)!
Also solved by Paolo Perfetti, Department of Mathematics, Tor Ver-gata University, Rome, Italy and Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain.
Editor’s Comment: On the last issue we forgot to acknowledge two more solu-tions of Problem 55 of the MathContest section we received. One from Angel Plaza, independently, and another from Yiri D. Valencia, Himar A. Fabelo (students), and Angel Plaza (jointly), University of Las Palmas, Gran Canaria, Spain.
MATHNOTES SECTION
Calculating the limits of some real
sequences
D.M.B˘atinet¸u-Giurgiu, Neculai Stanciu, Anastasios Kotronis
Abstract. In this note we present new methods to calculate the limits of some particular sequences and their generalizations that appeared in some problem solv-ing journals.
1. Main Results
For this section we assume that:
(1) {an}n≥1is a positive sequence such that lim
n→+∞an= a ∈ R ∗
+ and n→+∞lim n (an+1− an) = b ∈ R,
(2) f : R∗+ → R∗+ is a monotone nondecreasing function with a continuous derivative f0, and that
(3) {xn}n≥1is a positive sequence for which there exists a number t ∈ R such that lim n→+∞ xn+1 xnnt+1 = x ∈ R ∗ +.
With the above assumptions, we can deduce the following results:
Proposition 1. lim
n→+∞n (f (an+1) − f (an)) = bf 0(a).
Proof. For n ∈ N∗, on each interval [an, an+1], if an+16= an the function f satisfies the assumptions of Lagrange’s theorem, so there exists a real number ξn between an and an+1, such that f (an+1) − f (an) = (an+1− an) f0(ξn), (and this remains valid with ξn= an if an = an+1,) thus
n (f (an+1) − f (an)) = n (an+1− an) f0(ξn).
Now, on account of 1 and 2, we have limn→∞ξn= a, because limn→∞an= a, and f0 is continuous, therefore,
lim
n→+∞n (f (an+1) − f (an)) =n→+∞lim n (an+1− an) · limn→+∞f 0(ξ
n) = bf0(a),
which is the desired result.
Remark. Note that, we do not need the full strength of 2, we only need that f be a real valued function defined on (0, +∞) with a continuous derivative. This remark will help us in the next proposition.
Proposition 2. limn→+∞ f (a n+1) f (an) n = ebf 0 (a)f (a) .
Proof. According to the preceding remark, we can Apply Proposition 1, to the function ln(f ) to conclude that
lim
n→∞n (ln f (an+1) − ln f (an)) = b f0(a)
f (a).
Now, the Proposition follows using the continuity of the exponential function.
Proposition 3. lim n→+∞ nt+1 n √ xn = et+1 x .
Proof. Consider vn = nn(t+1)/xn. We have vn+1 vn =(1 + n) (n+1)(t+1)x n nn(t+1)x n+1 = xn xn+1 (1 + n)(n+1)(t+1) nn(t+1) =xnn t+1 xn+1 1 + 1 n 1+t 1 + 1 n n1+t
Thus, using 3, we have lim n→∞
vn+1 vn =
e1+t
x , and the proposition follows immediately from the well-known property [17, p. 46]:
lim n→∞ vn+1 vn = ` =⇒ lim n→∞ n √ vn= `,
for every positive sequence {vn}n≥1.
For the following propositions we set un=
n+1√xn+1 (n + 1)t · nt n √ xn for n ≥ 2. Proposition 4. lim n→+∞un= 1, n→+∞lim u n
n= e, and n→+∞lim n(un− 1) = 1. Proof. Keeping the notation of the previous proof we have √nvn= nt+1/√nxn, thus
un= 1 + 1 n √nv n n+1√vn+1 So, from Proposition 3, we have limn→∞un= 1. Also,
unn= 1 + 1 n n · vn vn+1 · n+1√v n+1
and limn→∞unn= e follows also from Proposition 3. Finally, since
n(un− 1) = ln(unn) · un− 1
ln un ,
we conclude immediately that lim
n→+∞n(un− 1) = 1. Proposition 5. If for n ≥ 2, Cn = n+1√x n+1 (n + 1)t − n √ xn nt , then lim n→+∞Cn = xe −(t+1)
Proof. We have Cn= n+1√xn+1 (n + 1)t − n √ xn nt = n √ xn nt+1 · n(un− 1), so on account of Propositions 3 and 4:
lim n→+∞Cn=n→+∞lim n √ xn nt+1 · limn→+∞n(un− 1) = xe −(t+1).
This concludes the proof of the Proposition.
Proposition 6. If for n ≥ 2, we set Bn = f (an+1) n+1√xn+1 (n + 1)t − f (an) n √ xn nt , then lim n→+∞Bn= x (f (a) + bf0(a)) et+1 . Proof. We have Bn = f (an) n √ xn nt f (an+1) f (an) un− 1 = f (an) n √ xn nt (tn− 1), where tn:= f (an+1) f (an) un. Now, since limn→+∞tn
4
== f (a)f (a) · 1 = 1, we obtain limn→+∞tln tn−1n = 1, and on account of Propositions 2, 4: lim n→+∞t n n=n→+∞lim f (an+1) f (an) n · lim n→+∞u n n= e bf 0 (a) f (a) e = e f (a)+bf 0 (a) f (a) . Writing Bn = f (an) n √ xn nt+1 · tn− 1 ln tn ln tnn we get lim n→+∞Bn= f (a) x et+1· 1 · ln e f (a)+bf 0 (a) f (a) =x (f (a) + bf 0(a)) et+1 .
Remark. Proposition 6 may also be proved as follows: We have Bn= f (an+1) n+1√xn+1 (n + 1)t − f (an) n √ xn nt = f (an+1) n+1√xn+1 (n + 1)t − f (an+1) n √ xn nt + f (an+1) n √ xn nt − f (an) n √ xn nt = f (an+1) n+1√x n+1 (n + 1)t − n √ xn nt + n √ xn nt+1 · n (f (an+1) − f (an)) and Propositions 1, 3 and 5 yield
lim n→+∞Bn= f (a) x et+1 + x et+1bf
0(a) = x (f (a) + bf0(a)) e−(t+1).
2. Applications
On the following we preserve the notation of the previous section. A1. Evaluate limn→+∞
n+1p(n + 1)! − √n n!.
Solution: With xn = n!, n ∈ N∗, ana sequence satisfying the assumptions in 1., and f : R∗+−→R∗+, f (x) = 1 we have a = lim n→+∞an, x =n→+∞lim xn+1 nxn = lim n→+∞ (n + 1)! n!n = 1, t = 0 and from Proposition 6,
Bn= n+1 p (n + 1)! − n √ n!, so lim n→+∞ n+1p(n + 1)! − √n n!= e−1.
Bn is known as Traian Lalescu’s sequence (see [5]). A2. Evaluate limn→+∞
n+1√n + 1n+1p(n + 1)! − √nn√nn!. Solution: With xn = n!, n ∈ N∗, an= n √ n, and f : R∗+−→R∗+, f (x) = x we have a = 1, x = lim n→+∞ xn+1 nxn = lim n→+∞ (n + 1)! n!n = 1, t = 0 and from Proposition 1:
b = lim n→+∞(an+1− an) n =n→+∞lim n+1√ n + 1 −√nn n = lim n→+∞ (n + 1) n+1√ n + 1 − n√nn − n+1√ n + 1 = 1 − 1 = 0.
Now, from Proposition 6,
Bn= n+1√ n + 1n+1p(n + 1)! − √nn√n n!, so lim n→+∞ n+1√ n + 1n+1p (n + 1)! − √nn√n n!= e−1. A3. Let {un}n≥1, {vn}n≥1be real positive sequences, and c ∈ R∗with limn→+∞nuc+1n+1u
n = u ∈ R∗+and limn→+∞ vn+1 ncvn = v ∈ R ∗ +. Calculate limn→+∞ n+1 qu n+1 vn+1 − n qu n vn .
(This application is a generalization of [1].) Solution: We have lim n→+∞ un+1vn nvn+1un = lim n→+∞ un+1 nc+1u n · lim n→+∞ ncv n vn+1 = u v ∈ R ∗ + and taking f : R∗ +−→R∗+, f (x) = 1, xn = uvnn, n ∈ N∗ and {an}n≥1 any sequence satisfying the assumptions in 1., we get
a = lim n→+∞an, x =n→+∞lim xn+1 nxn = u v ∈ R ∗ +, t = 0
and lim n→+∞ n+1r un+1 vn+1 −r un n vn = x e(f (a) + bf 0(a)) = x e(1 + b · 0) = x e = u ve.
For c = 2, with the notation in [1], it is un= bn, vn= an, n ∈ N∗, an, bn∈ R∗+ and limn→+∞ an+1 n2an = limn→+∞ bn+1 n3bn = a ∈ R∗+, so lim n→+∞ n+1 s bn+1 an+1 −r bn n an ! = a ae = e −1
and we have solved [1].
A4. Let f, g : R → R such that f (x) + g(x) = 1, x ∈ R and Bn(f, g) = nf (x) n+1p (n + 1)! g(x) −√nn! g(x) . Calculate limn→+∞Bn(f, g). Solution: Setting wn(x) = n+1√ (n+1)! n √ n! g(x) , we have lim n→+∞wn(x) = n→+∞lim n+1p(n + 1)! n + 1 ! · √nn n! · n + 1 n !g(x) = 1 e · e · 1 g(x) = 1 and limn→+∞wln wn(x)−1 n(x) = 1. Furthermore, lim n→+∞(wn(x)) n = lim n→+∞ n+1p(n + 1)! n √ n! !n!g(x) = lim n→+∞ (n + 1)! n! · 1 n+1p(n + 1)! !g(x) = lim n→+∞ n + 1 n+1p(n + 1)! !g(x) = eg(x).
Now we note that
Bn(f, g) = nf (x) √n n! g(x) (wn(x) − 1) = n n √ n! n !g(x) (wn(x) − 1) = n √ n! n !g(x) ·wn(x) − 1 ln wn(x) ln (wn(x)) n so lim n→+∞Bn(f, g) = 1 e g(x) · 1 · lneg(x)= g(x)e−g(x).
If f (x) = cos2x, g(x) = sin2x, x ∈ R, then with the notation in [2], Bn(f, g) = Ln= ncos 2x n+1p (n + 1)! sin2x −√n n! sin2x , so lim n→+∞Bn(f, g) = n→+∞lim Ln = sin 2x · e− sin2x
and the solution of [2]
follows.
Remark: With the methods presented above, one can solve [1] to [16] and many other problems from various math problem solving journals.
References
[1] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 24, Mathproblems Mathematical journal, Vol.1, Issue 4, 2011, p.33, http://mathproblems-ks.com/?wpfb_dl=4
[2] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 67, Mathproblems Mathematical journal, Vol.3, Issue 2, 2013, p.140, http://mathproblems-ks.com/?wpfb_dl=10
[3] Maria B˘atinet¸u-Giurgiu, On Lalescu Sequences, Octogon Mathematical magazine, Vol.13, No.1A, April 2005, pp.198-202,
[4] D.M.B˘atinet¸u-Giurgiu, S¸iruri Lalescu, Revista Matematic˘a duin Timi¸soara, Nr.1-2, 1985, pp.33-38,
[5] Trian Lalescu, Problem 579, Gazeta Matematic˘a, Vol.VI, 1900-1901, pp.33-38,
[6] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 692, The Pentagon, Vol.71, No.1, 2011, p.54, http://www.pentagon.kappamuepsilon.org/pentagon/Vol_71_Num_1_Fall_2011.pdf [7] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 5208, School Science and Mathematics
jour-nal, April, 2012, p.1, http://www.ssma.org/Websites/ssma/images/Problems%20Section/ April-2012.pdf
[8] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 43, Mathproblems Mathematical journal, Vol.2, Issue 3, 2012, p.91, http://mathproblems-ks.com/?wpfb_dl=7
[9] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 704, The Pentagon, Vol.71, No.2, 2012, p.42, http://www.pentagon.kappamuepsilon.org/pentagon/Vol_71_Num_2_Spring_2012.pdf [10] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 11676, The American Mathematical Monthly,
Vol.119, No.9, November 2012, p.801,
[11] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 3713, Crux Mathematicorum, Vol.38, No.2, Feb-ruary 2012, p.63,
[12] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 715, The Pentagon, Vol.72, No.1, 2012, p.44, http://www.pentagon.kappamuepsilon.org/pentagon/Vol_72_Num_1_Fall_2012.pdf [13] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 234, La Gazeta de la RSME, Vol.16, No.3, 2013,
p.502, http://www.rsme.es/gacetadigital/english/vernumero.php?id=92
[14] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 2414, Revista Escolar de la Olimpiada Iberoamer-icana de Matem´atica, No.49, 2013, p.502, http://www.oei.es/oim/revistaoim/numero49/ Probs241_245.pdf
[15] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem 3764, Crux Mathematicorum, Vol.38, No.7, Sep-tember 2012, p.285.
[16] D.M.B˘atinet¸u-Giurgiu, N.Stanciu, Problem W3, J´ozsef Wildt International Mathematical Competition, Edition XXIII, 2013, p.285, Octogon Mathematical magazine, Vol.21, No.1, April 2013, p.229, http://www.uni-miskolc.hu/~matsefi/Octogon/volumes/Wildt_2013_1. pdf
[17] W.J. Kaczor, M.T. Nowak Problems in Mathematical Analysis I, Real Numbers, Sequences and Series, A.M.S., 2000.
D.M.B˘atinet¸u-Giurgiu: Department of Mathematics, Matei Basarab National College, Bucharest, Romania,
Neculai Stanciu: Department of Mathematics, Geogre Emil Palade General School, Buz˘au, Romania,[email protected],
JUNIOR PROBLEMS
Solutions to the problems stated in this issue should arrive before June 15, 2014
Proposals
21. Proposed by Dorlir Ahmedi, University of Prishtina, Republic of Kosova If a, b, c > 0 and b+ca ≥ 3/2 then prove that
a b + c + b c + a+ c a + b≥ 2
22. Proposed by Callegari Emanuele, Math. Dept. “Tor Vergata” University, Rome, Italy. We have a rectangular chessboard 3 × 20 made of square boxes 1 × 1. We also have 30 dominoes of size 1 × 2 or 2 × 1 that we want to use to cover the chessboard. How many are the different ways to do that?
23. Proposed by Callegari Emanuele, Math. Dept. “Tor Vergata” University, Rome, Italy. Mary has 5 baskets each containing 97 colored balls numbered from 0 to 96. The color of the balls
in the first basket is blue, in the second is green and in the third is red. The color of the balls in the fourth and fifth baskets is white. Mary picks up a ball from each basket in such a way that the sum of the five numbers is 96. How many different configurations can occur to Mary?
24. Proposed by Stanescu Florin, Serban Cioculescu school, Gaesti, jud.Dambovita, Romania Prove that in a triangle ABC the following inequality holds:
27r 2p ≤ ra a + rb b + rc c ≤ p 2r
where ra, rb, rc are the lengths of the rays of the excircles, r is the radius of the circle inscribed in the triangle, and p the semiperimeter of the triangle
25. Proposed by Proposed by Neculai Stanciu, “George Emil Palade” School, Buzˇau, Romania and D.M. B˘atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania. Find all pairs of real numbers (x, y) such that
p
Solutions
16. Proposed by D.M. B˘atinet¸u–Giurgiu, “Matei Basarab” National College, Bucharest, Romania, Neculai Stanciu, “ George Emil Palade” School, Buzˇau, Ro-mania. Determine all real number x satisfying
1 x − 3√x + 2+ 1 x −√x = 2 x − 2√x
Solutions by Codreanu Ioan Viorel , Maramures, Romania, Daniel V˘acaru, Pite¸sti, Romania, Omran Kouba, Damascus, Syria and the proposers (independently). The existence conditions are
x > 0, x − 3√x + 2 6= 0, x −√x 6= 0, x − 2√x 6= 0 x > 0 ∧ x − 3√x + 2 = (√x − 2)(√x − 1) 6= 0 ⇐⇒ x 6= 4, x 6= 1
x > 0 ∧ x −√x 6= 0 ⇐⇒ x 6= 1, x − 2√x 6= 0 ⇐⇒ x 6= 4 Thus if x ∈ (0, +∞)\{1, 4}, the l.h.s. of the equation becomes
1 (√x − 2)(√x − 1) + 1 √ x(√x − 1) = √ x +√x − 2 √ x(√x − 1)(√x − 2) = = 2( √ x − 1) √ x(√x − 1)(√x − 2) = 2 √ x(√x − 2) = 2 x − 2√x which equals the r.h.s.
Two incorrect solutions have been submitted.
17. Proposed by D.M. B˘atinet¸u–Giurgiu, “ Matei Basarab;; National College, Bucharest, Romania, Neculai Stanciu,“ George Emil Palade” School, Buz˘au, Ro-mania. Prove that the acute triangle ABC is equilateral if and only if
tan2A sin2B + cos2C + tan2B sin2C + cos2A + tan2C sin2A + cos2B = 9
Solution by the authors. If A = B = C the triangle is equilateral and the equality trivially holds.
Suppose now that the equality holds.
9 = U =X cyc tan2A sin2B + cos2C =P 1 cyc(sin 2B + cos2C) X cyc (sin2B + cos2C) X cyc tan2A sin2B + cos2C =1 3 X cyc (sin2B + cos2C) X cyc tan2A sin2B + cos2C
and Cauchy–Schwartz’s inequality yields
9 = U ≥ 1
3(tan A + tan B + tan C) 2
that is 27 = 3U ≥ X cyc
tan A 2
. Since ABC is acute, the tangent is defined on
(0, π/2) and then the convexity of tan x, x ∈ (0, π/2), through Jensens’inequality gives 27 = 3U ≥ X cyc tan A !2 ≥ 3 tanA + B + C 3 2 = 3 tan2π 3 = 27
Equality holds if and only if A = B = C. The injectivity of the tangent in the given domain imposes A = B = C = π
3 that is the triangle is equilateral.
Also solved by Titu Zvonaru, Com˘anesti¸, Romania and Omran Kouba, Damascus, Syria.
18.Proposed by Ercole Suppa, teramo, Italy Let K be the symmedian point of 4ABC, let D = AK ∩ BC. Denote by P and Q the intersection points (different from A) of AB and AC with the circumcircles of triangles 4ADC and 4ABD respectively. Show that P Q is parallel to BC.
Solution by D.M. Bˇatinet¸u-Giurghiu, Bucharest, Romania, Neculai Stan-ciu, Buzˇau, Romania and Titu Zvonaru, Comˇane¸sti, Romania.
Using the power of the point B with respect to the circumcircle of triangle ADC we obtain BP · BA = BD · BC, so
BP BA =
BD · BC BA2
Similar from the power of the point C with respect to the circumcircle of triangle ABD we obtain
CQ CA =
CD · BC CA2 Since AD is the symmedian from A we have
BD DC =
AB2 AC2
From last three equation we have BPBA =CQCA, so P Q is parallel to BC.
Remark The equivalence BP BA = CQ CA ⇐⇒ BD DC = AB2 AC2
shows that the reverse is true, i.e. if P Q is parallel to BC then AD is the symmedian from A.
Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria, and the proposer.
19.Proposed by Armend Shabani, University of Prishtina, Republic of Kosova. Find all integer solutions of the equation 3x+ x4= 5x.
Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria The answer is x ∈ {0, 2}.
Indeed, if x is a negative integer then clearly 3x+ x4 > 3x > 5x and x is not a solution. Moreover, if x is a nonnegative integer solution, then x must be even since both 3xand 5xare odd integers in this case.
It is clear that x = 0 and x = 2 are solutions. Now, if x ≥ 4 then
5x− 3x> 6 75 x (a) 6 7 · 5 x> x4 (b)
Combining (a) and (b) we see that for x ≥ 4 we have 5x− 3x > x4 and x is not a solution to the proposed equation. Thus {0, 2} are the only solutions to this equation. Let us now prove (a) and (b):
(a) We have 53x ≥ 5 3 4 > 7, so that 5x− 3x= 6 75 x+ 5x− 7 × 3x 7 > 6 75 x. Which is (a). (b) Let an=67 ·5 n
n4. Clearly, a4> 1 and, for n ≥ 4, we have an+1 an = 5 n 1 + n 4 ≥ 5 4 1 + 4 4 =256 125 > 1. Thus an> 1 for n ≥ 4, which is equivalent to (b)
Also solved by Arber Igrishta, Mathematical Group Galaktika Shqiptare, Albania, Ioan Viorel Codreanu, Satulung, Maramures, Romania, D.M. Bˇatinet¸u-Giurghiu, Bucharest, Romania, Neculai Stanciu, Buzˇau and Titu Zvonaru, Comˇane¸sti, Romania, and the proposer.
20. Proposed by Paolo Perfetti, Math. Dept. “ Tor Vergata” University, Rome, Italy. Let a0= a1= a2= 1 and for n ≥ 1
an+2=
anan+1 an+ an−1 Find an for any n.
Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let xn = an/an+1, with this notation we have xn+1= 1 + xn−1. Hence x2n− x0= n X k=1 (x2k− x2k−2) = n x2n+1− x1= n X k=1 (x2k+1− x2k−1) = n
Thus, x2n= x2n+1= n + 1 for n ≥ 0. So, for n ≥ 1, we have a2n−2 a2n = x2n−2x2n−1= n2 Multiplying, we obtain a2n= 1 (n!)2, and a2n+1 = a2n xn = 1 n! · (n + 1)!. Finally an = 1 bn/2!c · dn/2!e,
ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 4, Issue 2 (2014), Pages 263-302
Editors: Valmir Krasniqi, Jos´e Luis D´ıaz-Barrero, Armend Sh. Shabani, Paolo Perfetti, Mohammed Aassila, Mih´aly Bencze, Valmir Bucaj, Emanuele Callegari, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba, Cristinel Mortici, Jozsef S´andor, Ercole Suppa, David R. Stone, Roberto Tauraso, Francisco Javier Garc´ıa Capit´an.
PROBLEMS AND SOLUTIONS
Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction. Proposals should be accompanied by solutions. An asterisk (*) indicates that nei-ther the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a com-mittee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]
Solutions to the problems stated in this issue should arrive before October 15, 2014
Problems
96. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. Let p≥ 1 be an integer and let x ∈ R. Prove that
∞ X n=1 np ex− 1 − x 1!− x2 2! − · · · − xn n! = ex Z x 0 Qp(t)dt,
where Qp is a polynomial of degree p which satisfies the equation Qp+1(x) = xQ0p(x) + xQp(x) with Q1(x) = x
97. Proposed by Sava Grozdev and Deko Dekov (Jointly), Bulgaria. Given points U and P in the plane of4ABC. Let UaUbUc be the cevian triangle of U . Denote by Ra, Rb and Rc the reflections of Ua, Ub and Uc in P , respectively. If the lines ARa, BRb and CRc concur in a point, we say that the Prasolov product of U and P is defined. In this case the intersection point of the lines is the Prasolov product of U and P. Note that if U is the orthocenter of 4ABC and P is the nine-point center of4ABC, then the Prasolov product is known as the Prasolov point. Prove
c
2010 Mathproblems, Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e. 263
that the Prasolov product is defined, provided U is the Nagel point of4ABC and P is the Spieker center of4ABC. The problem could be re-formulated as follow. Let U a be the point at which the A-excircle meets the side BC of 4ABC, and define U b and U c similarly. Let P be incenter of the medial triangle of 4ABC. Denote by Ra, Rb and Rc the reflections of Ua, Uband Uc in P , respectively. Prove that the lines ARa, BRb and CRc concur in a point.
98. Proposed by Anastasios Kotronis, Athens, Greece. Show that n! nn n X k=0 nk k! − +∞ X k=n+1 nk k! ! = 4 3+O(n −1).
99. Proposed by Li Yin, Department of Mathematics, Binzhou University, Binzhou City, Shandong Province, 256603, China. Calculate
∞ Y n=2 " 4 e2 1 + 1 n 2n+1 (n− 1)(n + 1) (2n− 1)(2n + 1) #
100. Proposed by D.M. B˘atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania, and Neculai Stanciu, “George Emil Palade” School, Buzˇau, Romania. Let (γn)n≥1 be the sequence defined by γn =− ln n +Pnk=1
1
k and let γ = lim
n→∞γn. Consider a continuous function f : (0, +∞) → (0, ∞). Find lim n→∞ n p (2n− 1)!! Z γn γ f (x)dx.
101. Proposed by Florin Stanescu, Serban Cioculescu school, city Gaesti, jud. Dambovita, Romania. Consider a real function f : [a, b]→ R (with a > 0,) having a positive and increasing derivative. Show that for every positive integer n with n≥ 2 the following inequality holds
Z b a f (x)dx≤ nn − 1 (b− a)(bnf (b)− anf (a)) bn− an − bf (b)− af(a) n .
102. Proposed by Marcel Chirit¸ˇa, Bucharest, Romania. Let f : (0, +∞) → R be a bounded continuous function. Suppose that the limit
lim x→∞x
α
|f(x + 2) − 2f(x + 1) + f(x)| = a ∈ R, exists for some α∈ [0, 1]. Find the value(s) of a.
Solutions
No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.
88.Proposed by Hun Min Park, Korea Advanced Institute of Science and Technol-ogy, Daejeon, South Korea. Suppose that three real numbers a, b, c(0≤ a, b, c, ≤ 1) satisfies the following equality;
X cyc a − b 1− ab· a 1− a2 = 0 Prove that a = b = c. (Note thatX cyc
means ’cyclic sum’X cyc
f (x, y, z) = f (x, y, z) + f (y, z, x) + f (z, x, y)). Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. It must be a, b, c6= 1. The equality is
X cyc 1 1− a2 = X cyc 1 1− ab which is, by virtue 0≤ a, b, c < 1,
X cyc ∞ X k=0 (a2)k =X cyc ∞ X k=0 (ak)2=X cyc ∞ X k=0 (ab)k We know that a2+ b2+ c2 ≥ ab + bc + ca, then (ak)2+ (bk)2+ (ck)2≥ (ab)k+ (bc)k+ (ca)k with the equality if and only if a = b = c. The result follows.
Solution 2 by Moshe Goldstein and Moti Levy, Rehovot, Israel. Without loss of generality, we may assume that a≥ b ≥ c.
Suppose that a = b, then 0 =X cyc a− b 1− ab a 1− a2 = a− c 1− ac a 1− a2+ c− a 1− ca c 1− c2 = (a− c)(1 (1 + ac) − ac) (1 − a2) (1− c2). It follows that a = c, which implies a = b = c. Now suppose, a6= b, that is a > b ≥ c. Then
a 1− a2 > b 1− b2 ≥ 0, (1) 1 1− ab > 1 1− bc≥ 0, (2) 1 1− ca > 1 1− bc≥ 0. (3)
It follows from (1) that X cyc a− b 1− ab a 1− a2 > c 1− c2 X cyc a− b 1− ab. (4)
Using (2) and (3) we obtain, X cyc a− b 1− ab > 1 1− bc X cyc (a− b) = 0. (5)
Inequalities (4) and (5) imply that Pcyc1a−ab−b 1−aa2 > 0, which contradicts the as-sumption in the problem statement, hence a = b and we are done.
Solution 3 by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. For a, b, c∈ [0, 1) we consider
F (a, b, c) = a− b 1− ab· a 1− a2 + b− c 1− bc· b 1− b2+ c− a 1− ca· c 1− c2 Noting that x− y 1− xy · x 1− x2 = (1− xy) − (1 − x2) (1− xy)(1 − x2) = 1 1− x2 − 1 1− xy = ∞ X n=1 (x2n− xnyn) we conclude that F (a, b, c) = ∞ X n=1 (a2n+ b2n+ c2n− anbn− bncn− cnan) = 1 2 ∞ X n=1 (an− bn)2+ (bn− cn)2+ (cn− an)2 ≥ 12(a− b)2+ (b − c)2+ (c − a)2 So, if F (a, b, c) = 0 then a = b = c.
Also solved by Arkady Alt, San Jose, California, USA; D.M. Bˇ atinet¸u-Giurgiu, “Matei Basarab” National College, Bucharest, Romania and Neculai Stanciu, “George Emil Palade” School, Buzˇau, Romania(Jointly); and the proposer.
89. Mohammed Aassila, Strasbourg, France.
Let S be the set of positive integers that does not contain the digit 7 in their decimal representation. Prove that
X n∈S
1
n < +∞.
Solution 1 by Henry Ricardo, New York Math Circle, New York, USA. We can write S = S1+ S2+ S3+· · · . where Si is the sum of all terms of the harmonic series whose denominators contain exactly i digits, all different from 7. Now the number of i-digit numbers that do not contain the digit 7 is 8·9i−1: There are 8 choices for the first digit, excluding 0 and 7, and 9 choices for the remaining i− 1 digits.
Furthermore, each number in Siis of the form 1/m, where m is an i-digit number. So m≥ 10i−1, which implies that 1/m≤ 1/10i−1. Therefore
S = ∞ X i=1 Si ≤ ∞ X i=1 8· 9i−1 10i−1 = 8 ∞ X i=1 9 10 i−1 = 80, so S converges by comparison.
Remark: This method can be used to show convergence of the sum of reciprocals of integers that do not contain the digit k∈ {0, 1, 2, . . . , 9}.
Solution 2 by Moti Levy, Rehovot, Israel. We partition the set S into disjoint subsets{Sk}, S = ∞ [ k=1 Sk,
where Sk := S∩10k−1, 10k. Define the sequence{ak}k≥1 by ak = X n∈Sk 1 n. Clearly,P∞k=1ak =Pn∈S n1.
Now we show by mathematical induction that the number of terms in Sk is less than 9k.
The number of terms in S1 is 8 < 91.
Suppose that it is true that the number of terms in Sk is less than 9k, then we have to show that the number of terms in Sk+1 is less than 9k+1. To show this, we split the set Sk+1 into nine intervals
Sk+1= 9 [ α=1
S∩α10k, α10k+ 1, . . . , (α + 1)10k.
The seventh intersection S∩7· 10k, 8· 10kis empty, of course.
The other intervals have the same number of terms, which is equal to the number of terms of the interval S∩1, 10k. By the induction hypothesis, it is less than 9 + 92+
· · · + 9k.
Hence, the number of terms in Sk+1is less than 8 9 + 92+· · · + 9k< 9k+1. Each term in Sk+1is not less than 10k, therefore ak+1=Pn∈Sk+1
1 n < 9k+1 10k. X n∈S 1 n= ∞ X k=1 ak < ∞ X k=1 9k 10k−1 = 90.
Reference: A. J. Kempner, A Curious Convergent Series, Amer. Math. Monthly, 21(2), (Feb. 1914) pp. 48-50.
Solution 3 by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia. Denote the number of terms of the given series between 1
10k and 10k+11 by nk for k≥ 0. n0= 8 and nk+1= 9nk, k≥ 0. Hence nk= 8· 9k.
Then X n∈S, n<10m+1 1 n < 1 +1 2 +· · · + 1 9 + 1 10+· · · + 1 99 +· · · + 1 10m +· · · + 1 99· · · 9 < 1· n0+ 1 10· n1+· · · + 1 10m · nm = 8 1 + 9 10+ 9 10 2 · · · + 9 10 m! < 8· 1 1− 9 10 = 80. ThereforePn∈S n1 < +∞.
Also solved by Roberto de la Cruz Moreno, Center of Recreational Math-ematics, Campus of Bellaterra, Barcelona, Spain; Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria; Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria; and the proposer.
90. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technol-ogy, Damascus, Syria. Let n be a positive integer, prove that
n X k=0 (−1)b√kc ≤ √ n and determine the cases of equality.
Solution 1 by Moti Levy, Rehovot, Israel. The sequence n(−1)b √
kco k≥0 is composed of consecutive runs of +1 or−1.
Let p be a positive integer. If (p− 1)2≤ n ≤ p2
−1 then d√ne = p and the sequence n
(−1)b √
kcon
k=0 is composed of p runs. The length of the m-th run is m2
− (m − 1)2= 2m− 1, for m = 1, . . . , p, and the sign of the units in the m-th run is (−1)m−1. The first run is{1}, the second run is{−1, −1, −1} and the third run is {1, 1, 1, 1, 1}, etc.
If (p− 1)2≤ n < p2 − 1 then n X k=0 (−1)b √ kc ≤ p−1 X m=1 (−1)m−1(2m− 1) ! + (−1)p−1n + 1− (p − 1)2 = (−1)p(p− 1) + (−1)p−1n + 1− (p − 1)2 = (−1)p(p− 1) −n + 1− (p − 1)2 = (p − 1) −n + 1− (p − 1)2 = p2− 1− n− p < p. If n = p2 − 1, n X k=0 (−1)b √ kc = p X m=1 (−1)m−1(2m− 1) = p = √ n, and these are the cases of equality.
Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Let q≤√k < q + 1. Thus
n X k=0 (−1)b√kc= bX√nc q=0 min{(q+1)2 −1,n} X k=q2 (−1)q = b√n+1X−1c q=0 (−1)q (q+1)2 −1 X k=q2 1 + bX√nc q=b√n+1−1c+1 (−1)q n X k=q2 1 = b√n+1−1cX q=0 (−1)q (q+1)2−1 X k=q2 1 + bX√nc q=b√n+1c (−1)q n X k=q2 1 If n = p2
− 1 it is b√n + 1c > b√nc and the second contribution is absent so for n = p2 − 1 we have b√n+1X−1c q=0 (−1)q (q+1)2 −1 X k=q2 1 = b√n+1X−1c q=0 (−1)q(2q + 1) = (−1)b√n+1−1c(b√n + 1− 1c + 1) = (−1)bp−1c(bpc) = (−1)bp−1cp Doing the modulus we get
b√n+1X−1c q=0 (−1)q (q+1)2 −1 X k=q2 1 = p =dpe = d √ ne Now let n be such that n = p2
− r where 2 ≤ r ≤ 2p − 1 in such a way that (p− 1)2 ≤ n ≤ p2 − 2. We have that √ n + 1 < 1 +b√nc. Indeed it yields p2− r + 1 < (1 + p − 1)2 ⇐⇒ r > 1
which evidently holds true. This implies thatb√n + 1c = b√nc. Thus we have bX√nc q=0 min{(q+1)2−1,n} X k=q2 (−1)q= b√n+1−1cX q=0 (−1)q (q+1)2−1 X k=q2 1 + bX√nc q=b√n+1−1c+1 (−1)q n X k=q2 1 = = b√n+1X−1c q=0 (−1)q(2q + 1) + ( −1)b√nc(n− (b√nc)2+ 1) = = (−1)b√n+1−1c(b√n + 1− 1c + 1) + (−1)b√nc(n− (b√nc)2+ 1) = = (−1)b√n+1−1cb√n + 1c + (−1)b√nc(n− (b√nc)2+ 1) = =−(−1)b√n+1cb√n + 1c + (−1)b√nc(n− (b√nc)2+ 1) Sinceb√n + 1c = b√nc, we have
−(−1)b√n+1cb√n + 1c + (−1)b√nc(n− (b√nc)2+ 1) ≤ ≤ −b√n + 1c + (n − (b√nc)2+ 1) =√n + 1− (n − (b√nc)2+ 1)≤ ≤√n + 1− 1
and the last step is to show √
n + 1− 1 ≤ d√ne Recall that n = p2− r thus we need to show
p
p2− r + 1 ≤ d√ne = 1 + b√nc = 1 + p − 1 which clearly holds.
Also solved by Arkady Alt, San Jose, California, USA; Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria; and the proposer.
91.Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania. Calculate Z 1 0 Z 1 0 ln(1− x) ln(1 − xy)dxdy.
Solution 1 by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. The answer is 3− 2ζ(3).
Let the considered integral be denoted by I. Since Z 1
0
ln(1− xy)dy =h−(1− xy)x ln(1− xy) − yiy=1 y=0=− (1− x) x ln(1− x) − 1 we see that I = Z 1 0 ln2(1− x) − ln(1 − x) −ln 2 (1− x) x dx = Z 1 0 ln2x− ln x − ln 2 x 1− x dx Now, noting that
3x− 3x ln x + x ln2x0= ln2x− ln x we see that Z 1 0 ln2x− ln xdx = 3 Also, since Z 1 0 xnln2xdx = Z ∞ 0 t2e−(n+1)tdt = Γ(3) (n + 1)3 = 2 (n + 1)3
we see that Z 1 0 ln2x 1− xdx = ∞ X n=0 Z 1 0 xnln2xdx = ∞ X n=0 2 (n + 1)3 = 2ζ(3) Finally I = 3− 2ζ(3).
Solution 2 by Anastasios Kotronis, Athens, Greece. It is easy to see that for k a positive integer:
X n≥1 1 n− 1 n + k = 1 +1 2 + 1 3+· · · + 1 k = Hk (1)
the k-th Harmonic Number.
Now, in what follows, the change of the way of summation and of summation-integration order, whenever it takes place, is justified by the constant sign of the summands-integrands. I : = Z 1 0 Z 1 0 ln(1− x) ln(1 − xy) dx dy = − Z 1 0 Z 1 0 X k≥1 (xy)k k ln(1− x) dx dy =− Z 1 0 X k≥1 yk k Z 1 0 xkln(1− x) dx dy = Z 1 0 X k≥1 yk k Z 1 0 xkX n≥1 xn n dx dy = Z 1 0 X k≥1 yk k X n≥1 1 n Z 1 0 xn+kdx dy = Z 1 0 X k≥1 X n≥1 yk nk(n + k + 1)dy =X k≥1 X n≥1 1 nk(n + k + 1) Z 1 0 ykdy =X k≥1 X n≥1 1 nk(k + 1)(n + k + 1) =X k≥1 X n≥1 1 k(k + 1)2n− 1 k(k + 1)2(n + k + 1) =X k≥1 1 k(k + 1)2 X n≥1 1 n− 1 (n + k + 1) =−X k≥1 1 k + 1− 1 k+ 1 (k + 1)2 X n≥1 1 n− 1 (n + k + 1) =−X k≥1 1 k + 1− 1 k X n≥1 1 n− 1 (n + k + 1) + 1−X k≥1 1 k2 X n≥1 1 n − 1 n + k : = A + 1− B
For A, from (1) and summing by parts we have:
A =−X k≥1 1 k + 1 − 1 k Hk+1=− Hk+1 k +∞ 1 + X k≥1 1 k + 1(Hk+2− Hk+1) = 3 2 + X k≥1 1 k + 1− 1 k + 2 = 2
