Problem 11724
(American Mathematical Monthly, Vol.120, August-September 2013]) Proposed by Andrew Cusumano (USA).
Let f (n) =Pn
k=1kk and let g(n) =Pn
k=1f (k). Find
n→∞lim
g(n + 2)
g(n + 1) −g(n + 1) g(n)
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will show that
n→∞lim
g(n + 2)
g(n + 1)−g(n + 1) g(n)
= e.
Since g(n) =Pn
k=1(n + 1 − k)kk =Pn
k=1k(n + 1 − k)n+1−k, it follows that g(n)
nn =
n
X
k=1
k nk−1
1 −k − 1 n
n+1−k
= 1 + 2 n
1 − 1
n
n−1
+ h(n) = 1 + 2e−1
n + o 1 n
.
because
0 ≤ h(n) =
n
X
k=3
k nk−1
1 − k − 1 n
n+1−k
≤ 3 n2+ 4
n3 + 1 n4
n
X
k=5
k ≤ 3 n2 + 4
n3 +n2+ n 2n4 .
Hence
g(n + 1) nn+1 =
1 + 1
n
n+1
· g(n + 1) (n + 1)n+1 = e
1 + 1
2n + o 1 n
1 + 2e−1
n + o 1 n
= e +e + 4 2n + o 1
n
and
g(n + 2) nn+2 =
1 + 2
n
n+2
· g(n + 2) (n + 2)n+2 = e2
1 + 2
n+ o 1 n
1 +2e−1
n + o 1 n
= e2+2e2+ 2e
n + o 1 n
.
Finally g(n + 2)
g(n + 1)−g(n + 1)
g(n) = g(n + 2)g(n) − g(n + 1)2 g(n + 1)g(n)
=n
e + o(1)
"
e2+2e2+ 2e
n + o 1 n
1 +2e−1
n + o 1 n
−
e +e + 4 2n + o 1
n
2#
=n
e + o(1) e2
n + o 1 n
= e + o 1 n
→ e.
