Problem 12079
(American Mathematical Monthly, Vol.125, December 2018) Proposed by M. Omarjee (France).
Choosex1 in(0, 1), and let for n ≥ 1,
xn+1= 1 n
Xn k=1
ln(1 + xk).
Compute lim
n→∞xnln(n).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We first note that for n ≥ 1, 0 < xn+1= log(1 + xn)
n +(n − 1)xn n < xn
n +(n − 1)xn
n = xn ≤x1,
which means that the sequence is positive and strictly decreasing. Therefore it has a limit L ∈ [0, 1).
Moreover
n(xn−xn+1) = xn−log(1 + xn) and since for t > 0,
t2 2 −t3
3 < t − log(1 + t) < t2 2, we have that
1 6 =1
2 −1 3 <1
2 −xn
3 < n(xn−xn+1) x2n
< 1 2. Then
x2n
6n < xn−xn+1 =⇒
XN n=1
x2n
n < 6 XN n=1
(xn−xn+1) = 6(x1−xN+1) < 6
which implies that L = 0 otherwise we have a contradiction: the bounded partial sum on the left diverges as n → +∞ . Therefore, by the above inequality, xn →0 implies
n→∞lim
n(xn−xn+1) x2n
=1 2. Finally, by StolzCesaro theorem, we find the desired limit
n→∞lim xnln(n) = lim
n→∞
ln(n) 1/xn
SC= lim
n→∞
ln(n + 1) − ln(n) 1/xn+1−1/xn
= limn→∞ ln(1 + 1/n) − 1 1/n
| {z }
→1
· x2n
n(xn−xn+1)
| {z }
→2
· log(1 + xn) nxn
| {z }
→0
+n − 1
| {z }n
→1
) = 2.
