Problem 11935
(American Mathematical Monthly, Vol.123, October 2016)
Proposed by D. M. B˘atineu-Giurgiu (Romania), Anastasios Kotronis (Greece), and Neculai Stanciu (Romania).
Letf be a function from Z+ to R+ such thatlimn→∞f (n)/n = a, where a > 0. Find
n→∞lim
n+1
v u u t
n+1
Y
k=1
f (k) − n v u u t
n
Y
k=1
f (k)
.
Solution proposed by Moubinool Omarjee, Lyc´ee Henri IV, Paris, France, and Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, Italy.
Solution. Let
an= n v u u t
n
Y
k=1
f (k)
and let xn= (an/n)n. Then by Stolz-Cesaro,
n→∞lim lnan
n
= lim
n→∞
Pn
k=1ln f (k) − n ln n n
SC= lim
n→∞(ln(f (n + 1)) − (n + 1) ln(n + 1) + n ln n)
= lim
n→∞
ln f (n + 1) n + 1
− n ln
1 + 1
n
= lna e
.
Hence ann = √xn n→ ae, which implies that xn+1xn → ae. Moreover an+1
an → 1, and an+1
an
n
= (n + 1)nx
n n+1
n+1
nxn
=
1 + 1
n
n
·xn+1
xn · 1
n+1√xn+1 → e.
Therefore
n+1
v u u t
n+1
Y
k=1
f (k) − n v u u t
n
Y
k=1
f (k) = an+1− an= an
n ·
a
n+1
an − 1
ln
1 +a
n+1
an − 1
· ln an+1
an
n
→ a e.
