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 Xn 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 +Pn
k=1 1
k and let γ = lim
n→∞γn. Consider a continuous function f : (0, +∞) → (0, ∞). Find
n→∞lim pn
(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≤ n n− 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
xlim→∞xα|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 + ac)
(1− 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 P
cyc a−b 1−ab
a
1−a2 > 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
≥ 1
2(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 =P
n∈S 1 n.
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· 10k
is 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=P
n∈Sk+1
1
n <910k+1k. 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 101k and 10k+11 by nk for k≥ 0. n0= 8 and nk+1= 9nk, k≥ 0. Hence nk= 8· 9k.
Then
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
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 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 Doing the modulus we get
which evidently holds true. This implies thatb√
n + 1c = b√nc. Thus 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
pp2− 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) −ln2(1− x) x
dx
= Z 1
0
ln2x− ln x − ln2x 1− x
dx Now, noting that
3x− 3x ln x + x ln2x0
= ln2x− ln x we see that
Z 1 0
ln2x− ln x dx = 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
Solution 2 by Anastasios Kotronis, Athens, Greece. It is easy to see that for k a positive integer:
X
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.
For A, from (1) and summing by parts we have:
A =−X
For B, from (1), where ζ is the Riemann zeta function.
So B = 2B− 2ζ(3) and hence B = 2ζ(3). Finally I = 3 − 2ζ(3) ≈ 0.595886.
Solver’s Note: A reference to the details presented here is “ M. S. Klamkin,Amer.
Math. Monthly, 59 (1952) pp. 471–472”. See also page 6 of the Collected Contri-butions of M. S. Klamkin to the Amer. Math. Monthly.
Solution 3 by Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria. For n∈ N>0, we can integrate by parts to get :
Z 1
Where Hn is the n-th harmonic number, therefore I = Now, we can simplify the latter series to be :
A = The first sum is easy to calculate :
Xn
Since Hn ∼ ln(n) and ln(n) = o(n) we get : A =
X∞ k=1
Hk+1
k −Hk+1
k + 1 = lim
n→+∞2− 1
n + 1−Hn+1
n + 1 = 2.
To calculate the other sum we use the following result for m > 0:
Z 1 0
xmln2x dxx=e−t/(m+1)= 1 (m + 1)3
Z ∞ 0
t2e−tdt = 2
(m + 1)3 (2) From (1) we get :
B = X∞ k=1
Hk
k2 =− X∞ k=1
Z 1 0
tk−1
k ln(1− t) dt = Z 1
0
ln2(1− t) t dt =
Z 1 0
ln2t 1− t dt.
Using the result (2) we get B =
X∞ k=0
tkln2t dt = X∞ k=0
2
(k + 1)3 = 2ζ(3).
Now, it is clear that :
I = A− (B − 1) = 3 − 2ζ(3).
Also solved by Arkady Alt, San Jose, California, USA; AN-anduud Prob-lem Solving Group, Ulaanbaatar, Mongolia; B.C. Greubel, Newport News, VA, USA; Moti Levy, Rehovot, Israel; Paolo Perfetti, Depart-ment of Mathematics, Tor Vergata University, Rome, Italy; and the proposer.
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≥0 be 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≥0 is the sequence of Lucas numbers. Prove that bn is not a perfect square, for every n∈ N \ {1}.
Solution 1 by Moti Levy, Rehovot, Israel. By its definition, bn is a perfect square if and only if L2n is a perfect square.
It was proved in the article, J. H. E. Cohn, ”Square Fibonacci Numbers, Etc.”
Fibonacci Quarterly 2 1964, pp. 109-113, that if Lkis perfect square then k = 1 or k = 3.
It follows that bn is not a perfect square for every n∈ N.
Solution 2 by Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria. Let mod(m, n) be the remainder on division of m by n.
With a quick evaluation we have
{mod(n2, 5)|n ∈ N} = {0, 1, 4, 4, 1, 0, 1, ...} = {0, 1, 4}
This is true because of the periodicity of the mod sequence : mod(n + 5k, 5) = mod(n, 5) for any n, k∈ N.
Hence mod(a2n, 5)∈ {1, 4} for all n ∈ N.
Let’s take a look on mod(Ln, 5), Here’s a lemma :
mod(Ln, 5) =
2 if mod(n, 4) = 0;
1 if mod(n, 4) = 1;
3 if mod(n, 4) = 2;
4 if mod(n, 4) = 3.
(lemma 1)
Proof : for n ∈ {0, 1, 2, 3} it is clearly true. suppose it is true for some integers 4k, 4k + 1, 4k + 2, 4 + 3, then
mod(L4k+4, 5) = mod(L4k+3+ L4k+2, 5) = mod(4 + 3, 5) = 2 mod(L4k+5, 5) = mod(L4k+4+ L4k+3, 5) = mod(2 + 4, 5) = 1 mod(L4k+6, 5) = mod(L4k+5+ L4k+4, 5) = mod(1 + 2, 5) = 3 mod(L4k+7, 5) = mod(L4k+6+ L4k+5, 5) = mod(3 + 1, 5) = 4 Which means that (lemma 1) is true for any integer n by strong induction.
For n = 0, we have mod(b0, 5)∈ {2, 3}, which means that b0∈ {0, 1, 4}, then b/ 0 is not a perfect square.
For n > 1, we have 2n = 4k where k ∈ N>0, therefore : mod(L2n, 5) = 2, which implies that
mod(bn, 5) = mod(2a2n, 5)∈ {2, 3}
Which means that bn∈ {0, 1, 4}, then b/ n is not a perfect square for n > 1.
Conclusion : bn is not a perfect square for any n∈ N\{1}.
Also solved by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria; and the proposer.
93.Proposed by Anastasios Kotronis, Athens, Greece. For x∈ (−1, 1), evaluate
+∞
X
n=1
(−1)n+1n
tan−1x− x +x3
3 − · · · + (−1)n+1x2n+1 2n + 1
.
Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.
Let
fn(x) = tan−1x− x +x3
3 − · · · + (−1)n+1x2n+1 2n + 1 Clearly we have
fn0(x) = 1 1 + x2 −
X2n k=0
(−x2)k = 1
1 + x2 −1− (−x2)n+1
1 + x2 = (−1)n+1x2n+2 1 + x2 Thus
fn(x) = (−1)n+1 Z x
0
t2n+2 1 + t2dt
It follows that
Finally, noting that t4
we conclude that X∞
which is the desired conclusion.
Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. Clearly
X∞ Since|x| < 1, we can rearrange the series as
X∞
X∞ By summing up the three contributions we obtain
−1
Solution 3 by Arkady Alt, San Jose, California, USA.
Let S (x) :=
Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia; Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria; Moti Levy, Rehovot, Israel; and the proposer.
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.
Solution 1 by Moti Levy, Rehovot, Israel.
Lemma 1. Let −→r1, −→r2, . . . , −→rnbe n arbitrary points on a plane. Let w1, w2, . . . , wn
be n positive weights, such that Pn
k=1wk = 1. Let −→c denote the weighted average vector defined by
−
→c :=
Xn m=1
wm−→rm. Then the following holds:
Xn m=1
wm|−→z − −→rm|2= Xn m=1
wm|−→c − −→rm|2+|−→c − −→z|2
Proof.
wm|−→z − −→rm|2= wm(−→z − −→rm)· (−→z − −→rm) = wm
|−→z|2+|−→rm|2− 2−→z · −→rm
wm|−→c − −→rm|2= wm(−→c − −→rm)· (−→c − −→rm) = wm
|−→c|2+|−→rm|2− 2−→c · −→rm
wm|−→z − −→rm|2− wm|−→c − −→rm|2= wm
|−→z|2− |−→c|2
+ 2wm(−→c − −→z )· −→rm
Summing,
Xn m=1
wm|−→z − −→rm|2− Xn m=1
wm|−→c − −→rm|2
=
|−→z|2− |−→c|2Xn
m=1
wm+ 2 (−→c − −→z )· Xn m=1
wm−→rm
=|−→z|2− |−→c|2+ 2 (−→c − −→z )· −→c
=|−→z|2− |−→c|2+ 2|−→c|2− 2−→z · −→c
=|−→z|2− 2−→z · −→c +|−→c|2=|−→c − −→z|2.
This proves the lemma.
It follows from the lemma that the vector which minimizes the weighted sum Pn
m=1wm|−→z − −→rm|2 is −→c .
Now we apply the lemma to our problem:
Let −→rA,−→rBand −→rCbe vectors from the origin to the triangle vertices, and let the weights be:
wA=
1 b2 1
a2 +b12 +c12
= c2a2 a2b2+ a2c2+ b2c2 wB= a2b2
a2b2+ a2c2+ b2c2 wC= b2c2
a2b2+ a2c2+ b2c2, then the original problem becomes:
Find a vector −→rP which minimizes the weighted sum
wA|−→rP− −→rA|2+ wB|−→rP − −→rB|2+ wC|−→rP − −→rC|2. The answer is
−
→rP = wA−→rA+ wB−→rB+ wC−→rC. If the triangle is equilateral then the point P is the centroid.
Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. We will use the following Lemma:
Lemma 2. Consider n points A1, . . . , An in the plane P, and let λ1, . . . , λn be n positive numbers. We consider, the real function f :P → R defined by
f (M ) = Xn k=1
λk|AkM|2
Let also G be the barycenter of the wighted points ((Ak; λk))1≤k≤n, that is G is the unique point G defined by
Xn k=1
λk−−→GAk = ~0,
then G is the unique point in the planeP where f attains its minimum.
Proof. Indeed,
f (M )− f(G) = Xn k=1
λk−−−→AkM2−−−→AkG2
= Xn k=1
λk−−−→AkM−−−→AkG
·−−−→AkM +−−→AkG
= Xn k=1
λk−−→GM·−−→GM + 2−−→AkG
= Xn k=1
λk
!
|GM|2+ 2−−→GM· Xn k=1
λk−−→AkG
= Xn k=1
λk
!
|GM|2.
This shows that f (M )≥ f(G) with equality if and only if M = G, and the lemma
follows.
In the proposed problem we have n = 3, A1= A, A2= B, A3= C and λ1= 1/b2, λ1 = 1/c2 and λ1 = 1/a2. Thus the desired minimum is attained at the unique point P which is the barycenter of the weighted points (A;b12), (B;c12), and (C;a12).
This point is the first Brocard point Ω, it is the unique point inside ABC such that
∠ΩAB = ∠ΩBC = ∠ΩCA.
(when the triangle ABC is labeled in counterclockwise order.) Also solved by the proposer.
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
where Wn :=(2n(2n)!!−1)!! is Wallis product.
Solution 1 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. We use
ln(1− x) = −x −1
2x2+ o(x2), ex= 1 + o(1), n!∼ (n/e)n√
2πn(1 + o(1)) We get
√1π
1−ln n 2n
n
= 1
√πexp
n ln
1−ln n 2n
= 1
√πexp
n
−ln n 2n + o(1
n)
= 1
√πexp
−ln n 2 + o(1)
= 1
√π√n(1 + o(1))
Wn= (2n)!
22n(n!)2 =(2n)2ne−2n√ 2π√
2n(1 + o(1)) 22nn2ne−2n2πn = 1
√π√n(1 + o(1))
It follows
n→∞lim
Wn
√1
π 1−ln n2n
n = 1
Solution 2 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. On one hand, using Stirling’s formula, we have
Wn= (2n)!
22n(n!)2 ∼ 2√πn(2n)2ne−2n 22n(2πn)n2ne−2n = 1
√πn, (1)
and on the other hand
1−ln n 2n
n
= exp
n ln
1−ln n 2n
= exp
n
−ln n 2n +O
ln2n n2
= exp
−ln n 2 +O
ln2n n
= 1
√nexp
O
ln2n n
= 1
√n
1 +O
ln2n n
Thus 1
√π
1−ln n 2n
n
∼ 1
√πn, (2)
and the desired conclusion follows from (1) and (2).
Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia; Haroun Meghaichi, University of Science and Techonology, Houari Boumediene, Algeria.