HOMEWORK 5 SOLUTIONS
Problem 1. (2008 B2) Let F0(x) = ln x. For n ≥ 0 and x > 0, let Fn+1(x) =Rx
0 Fn(t)dt. Evaluate
n→∞lim
n!Fn(1) ln n . Solution:
By directly computing Fn(x) for small n’s, we obtain the following conjectural identity
Fn(x) = xn
n! ln x −
n
X
k=1
1 k
!
. (1)
For convenience, let us define Gn(x) = xn
n! ln x −
n
X
k=1
1 k
!
for n ≥ 0 and x > 0, so that (1) can be stated as Fn(x) = Gn(x). For n ≥ 0 and x > 0 we note that
G0n(x) = xn−1
(n − 1)! ln x −
n
X
k=1
1 k
! +xn
n! · 1 x
= xn−1
(n − 1)! ln x −
n
X
k=1
1 k+1
n
!
= xn−1
(n − 1)! ln x −
n−1
X
k=1
1 k
!
= Gn−1(x). (2)
We proceed by induction on n to prove (1), i.e. Fn(x) = Gn(x). For n = 0, we immediately verify that G0(x) = ln x = F0(x). For the inductive step, we assume that (1) is true for n = m. This means that Fm(x) = Gm(x), while (2) says that Gmx = G0m+1(x). Now we calculate
Fm+1(x) = Z x
0
Fm(t)dt = Z x
0
Gm(t)dt = lim
r→0
Z x r
Gm(t)dt = lim
r→0
Z x r
G0m+1(t)dt
= lim
r→0(Gm+1(x) − Gm+1(r)) (by the Fundamental Theorem of Calculus)
= Gm+1(x) − lim
r→0Gm+1(r)
In order to complete the inductive step, it remains to show that lim
r→0Gm+1(r) = 0. This can be seen by writing Gm+1(r) = rm
(m + 1)!· ln r −Pm+1 k=1
1/r . Note that as r goes to 0, the fisrt factor clearly converges to 0 whereas the second factor also converges to 0 by L’Hopital’s rule.
Now (1) says that n!Fn(1) = −
n
X
k=1
1
k. Since f (x) = 1
x is strictly decreasing for x > 0, we obtain the following inequalities:
n
X
k=1
1 k = 1 +
n
X
k=2
1 k ≤ 1 +
n
X
k=2
Z k k−1
dx x = 1 +
Z n 1
dx
x = 1 + ln n,
n
X
k=1
1 k ≥
n
X
k=1
Z k+1 k
dx x =
Z n+1 1
dx
x = ln(n + 1) > ln n.
Hence we have that −1 + ln n
ln n ≤ n!Fn(1)
ln n < −ln n
ln n = −1. Since lim
n→∞
−1 + ln n ln n
= −1, we deduce that
lim
n→∞
n!Fn(1) ln n = −1.
Problem 2. (2011 A2) Let a1, a2, · · · and b1, b2, · · · be sequences of positive real numbers such that a1 = b1= 1 and bn = bn−1an− 2 for n = 2, 3, · · · . Assume that the sequence (bj) is bounded. Prove that
S =
∞
X
n=1
1 a1· · · an
converges, and evaluate S.
Solution:
The relation bn= bn−1an− 2 for n = 2, 3, · · · can be written as an= bn+ 2
bn−1 . Then for n = 2, 3, · · · , 1
a1· · · an
= 1 a1
1 a2· · · 1
an = 1 a1
b1b2· · · bn−1
(b2+ 2)(b3+ 2) · · · (bn+ 2)
= 3
b1+ 2
b1b2· · · bn−1
(b2+ 2)(b3+ 2) · · · (bn+ 2) (since a1= b1= 1)
= 3b1b2· · · bn−1
(b1+ 2)(b2+ 2) · · · (bn+ 2)
= 3 2
b1b2· · · bn−1(bn+ 2)
(b1+ 2)(b2+ 2) · · · (bn+ 2) − b1b2· · · bn−1bn
(b1+ 2)(b2+ 2) · · · (bn+ 2)
= 3 2
b1b2· · · bn−1
(b1+ 2)(b2+ 2) · · · (bn−1+ 2)− b1b2· · · bn−1bn (b1+ 2)(b2+ 2) · · · (bn+ 2)
For convenience, set cn = b1b2· · · bn−1bn
(b1+ 2)(b2+ 2) · · · (bn+ 2) for n = 1, 2, · · · , so that the above identity can be written as 1
a1· · · an
= 3
2(cn−1− cn) for n = 2, 3, · · · . Then for every positive integer m, we have
m
X
n=1
1 a1a2· · · an
= 1 a1
+
m
X
n=2
1 a1a2· · · an
= 1 +3 2
m
X
n=2
(cn−1− cn) = 1 +3
2(c1− cm) = 3
2(1 − cm), (3)
where the last identity follows by the fact that c1= 1 3.
Note that cm is clearly positive by construction since b1, b2, · · · are all positive. We also know that the sequence (bj) is bounded by some number M > 0. Now observe that
0 < cm= b1
b1+ 2· b2
b2+ 2· · · bm
bm+ 2 = 1 1 + 2/b1
· 1
1 + 2/b2
· · · 1 1 + 2/bm
<
1
1 + 2/M
m .
Since 1
1 + 2/M > 1, this implies that lim
m→∞cm = 0. Thus we deduce from (3) that S =
∞
X
n=1
1 a1· · · an
converges to 3 2.
Problem 3. (2006 A3) Let 1, 2, 3, · · · , 2005, 2006, 2007, 2009, 2012, 2016, · · · be a sequence defined by xk = k for k = 1, 2, · · · , 2006 and xk+1= xk+ xk−2005 for k ≥ 2006. Show that the sequence has 2005 consecutive terms each divisible by 2006.
Solution:
Define a sequence (yk) by yk = 0 for k = 1, 2, · · · , 2005, y2006= 1 and yk+1= yk+ yk−2005for k ≥ 2006.
Then we can easily compute y2006 = y2007 = · · · = y4011 = 1, y4012 = 2, y4013 = 3, · · · , y6017 = 2006. In particular, we have that y4011 = x1, y4012 = x2, · · · , y6017 = x2006. Now an easy (strong) induction shows that xk= yk+4010for k = 1, 2, · · · .
Note that (yk) is clearly a sequence of integers, so Useful Facts (2) from Lecture 5 (see the remark below) says that the sequence is periodic modulo 2006. In particular, since y1, y2, · · · , y2005are all zero and therefore divisible by 2006, there exist infinitely many integers k such that yk+1, yk+2, · · · , yk+2005are all divisible by 2006. Choose such a k with k ≥ 4011, then xk−4009 = yk+1, xk−4008 = yk+2, · · · , xk−2005= yk+2005 are all divisible by 2006.
Remark. For the sake of completeness, we state and prove Useful Facts (2) from Lecture 5.
Lemma. If (xn) is a sequence of integers satisfying a linear recurrence xn= a1xn−1+a2xn−2+· · ·+akxn−k, then (xn) is a periodic sequence in modulo m for every integer m.
Proof. Without loss of generality, we may assume that ak6= 0.
Consider the vectors of the form (xn, xn+1, · · · , xn+k−1). Note that there are only finitely many such vectors modulo m, so there exist integers p and N such that
(xN +p, xN +1+p, · · · , xN +k−1+p) ≡ (xN, xN +1, · · · , xN +k−1) (mod m). (4)
We claim that xn+p≡ xn(mod m) for all n. It suffices to prove that the following holds for all n:
(xn+p, xn+1+p, · · · , xn+k−1+p) ≡ (xn, xn+1, · · · , xn+k−1) (mod m). (5)
We proceed by the two-way induction on n. The case when n = N is given by (4). For the inductive step, we assume that (5) holds for n = r. We wish to prove that (5) holds for n = r − 1 and n = r + 1. For n = r + 1, it suffices to prove that xr+k+p≡ xr+k(mod m). This can be seen by
xr+k+p= a1xr+k−1+p+ a2xr+k−2+p+ · · · + akxr+p (by recursion)
≡ a1xr+k−1+ a2xr+k−2+ · · · + akxr(mod m) (by the induction hypothesis)
= xr+k (by recursion)
For n = r − 1, it suffices to prove that xr−1+p≡ xr−1(mod m). This can be seen by
xr−1+p= a−1k (xr+k−1+p− a1xr+k−2+p− · · · − ak−1xr+p) (by recursion)
≡ a−1k (xr+k−1− a1xr+k−2− · · · − ak−1xr) (mod m) (by the induction hypothesis)
= xr−1 (by recursion)
Problem 4. (2007 B3) Let x0 = 1 and for n ≥ 0, let xn+1= 3xn+ bxn
√5c. In particular, x1 = 5, x2=
26, x3= 136, x4= 712. Find a closed-form expression for x2007. (bac means the largest integer ≤ a. )
Solution:
Note that xn is an integer for n = 0, 1, 2, · · · .
By definition of the floor function b·c, we have the following inequality for n ≥ 0:
xn√
5 − 1 < bxn√
5c ≤ xn√
5. (6)
Then the recursion xn+1 = 3xn + bxn
√
5c gives an inequality (3 +√
5)xn − 1 < xn+1 ≤ (3 +√ 5)xn. Multiplying by 3 −√
5 yields
4xn− (3 −√
5) < (3 −√
5)xn+1≤ 4xn. (7)
Now observe that for n = 1, 2, · · · ,
xn+1− 6xn= −3xn+ bxn
√
5c ≤ −(3 −√
5)xn< −4xn−1+ (3 −√
5) < −4xn−1+ 1, xn+1− 6xn= −3xn+ bxn
√
5c > −(3 −√
5)xn− 1 ≥ −4xn−1− 1
where in each line, the first and the second inequality respectively follow from (6) and (7). Hence we have that −4xn−1− 1 < xn+1− 6xn < −4xn−1+ 1 for n = 1, 2, · · · . Since xn+1− 6xn is an integer, the only possibility is that xn+1− 6xn= −4xn. In other words, we have the following recurrence for n ≥ 1:
xn+1= 6xn− 4xn−1. (8)
The characteristic polynomial x2− 6x + 4 has two roots 3 ±√
5, so there exist constants c and d such that
xn = c(3 +√
5)n+ d(3 −√
5)n for n = 0, 1, 2, · · · .
Substituiting n = 0 and n = 1, we obtain a system of equations 1 = c + d and 5 = c(3 +√
5) + d(3 −√ 5), which can be solved by c = 5 + 2√
2
10 , d = 5 − 2√ 2
10 . Hence (8) yields the following expression of x2007: x2007= 5 + 2√
5 10 (3 +√
5)2007+5 − 2√ 5 10 (3 −√
5)2007.
