Problem 12143
(American Mathematical Monthly, Vol.126, November 2019) Proposed by J. A. Scott (UK).
Compute
n→∞lim
n
X
k=1
k n
k
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We show that the limit exists and it is equal to e e −1. Lower bound. Since
k n
k
=
1 + n − k k
−k
≥e−(n−k)
then n
X
k=1
k n
k
≥
n
X
k=1
en−k=
n−1
X
j=1
e−j. and
lim inf
n→+∞
n
X
k=1
k n
k
≥
∞
X
j=0
e−j= e
e −1. (1)
Upper bound. Let t ∈ (0, 1). Then we split the given sum as
n
X
k=1
k n
k
= X
1≤k≤tn
k n
k
+ X
tn<k<n−n1/3
k n
k
+ X
n−n1/3≤k≤n
k n
k
It follows that
X
1≤k≤tn
k n
k
≤ X
1≤k≤tn
tk ≤
∞
X
k=1
tk = t 1 − t. Moreover
X
tn<k<n−n1/3
k n
k
≤ X
t<k/n<1−1/n2/3
k n
tn
≤n
Z 1−1/n2/3 0
xtndx= n tn+ 1
1 − 1
n2/3
tn+1
and X
n−n1/3≤k≤n
k n
k
= X
0≤j≤n1/3
1 − j
n
n−j
= X
0≤j≤n1/3
exp
(n − j) ln
1 − j
n
≤ X
0≤j≤n1/3
exp
(n − j)
−j n
= X
0≤j≤n1/3
e−j·ej2/n≤e1/n1/3
n−1
X
j=0
e−j.
Therefore
n
X
k=1
k n
k
≤ t
1 − t+ n tn+ 1
1 − 1
n2/3
tn+1
+ e1/n1/3
n−1
X
j=0
e−j and
lim sup
n→+∞
n
X
k=1
k n
k
≤ t
1 − t+ 0 + e e −1. By letting t → 0+, we find
lim sup
n→+∞
n
X
k=1
k n
k
≤ e
e −1. (2)
Finally, by (1) and (2), we may conclude that our initial claim holds.
