Problem 11206
(American Mathematical Monthly, Vol.113, February 2006) Proposed by M. Ivan and A. Lupa¸s (Romania).
Find
n→∞lim 1 n
n
X
k=1
nn k
o2
where{x} = x − ⌊x⌋ denotes the fractional part of x.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We first note that
1 n
n
X
k=1
nn k
o2
=
n
X
k=1
1 k/n
2
·1 n is a Riemann sum corresponding to the integral
Z 1 0
1 x
2
dx.
Since n ≤ 1/x < n + 1 implies that
1 x
= 1 x− 1
x
= 1 x− n then
Z 1 0
1 x
2
dx = lim
N →∞
N
X
n=1
Z n1 1 n+1
1 x− n
2
dx
Now since Hn= log(n) + γ + o(1) and n! = nne−n√
2πn(1 + o(1)) then
N
X
n=1
Z 1n 1 n+1
1 x− n
2
dx =
N
X
n=1
Z n1 1 n+1
1 x2 −2n
x + n2
dx
=
N
X
n=1
−1
x− 2n log(x) + n2x
n1 1 n+1
=
N
X
n=1
1 + 2n log(n) − 2n log(n + 1) + 1 − 1 n + 1
=
N
X
n=1
2 + 2 (n log(n) − (n + 1) log(n + 1)) + 2 log(n + 1) − 1 n + 1
= 2N − 2(N + 1) log(N + 1) + 2 log((N + 1)!) + 1 − HN+1
= 2N + 2 log((N + 1)!/(N + 1)N+1) + 1 − log(N + 1) − γ + o(1)
= 2N + 2 log(e−(N +1)p2π(N + 1)(1 + o(1))) + 1 − log(N + 1) − γ + o(1)
= 2N − 2(N + 1) + log(2π) + log(N + 1) + 1 − log(N + 1) − γ + o(1)
= log(2π) − γ − 1 + o(1).
Therefore
n→∞lim 1 n
n
X
k=1
nn k
o2
= Z 1
0
1 x
2
dx = lim
N →∞
N
X
n=1
Z 1n 1 n+1
1 x− n
2
dx = log(2π) − γ − 1 ≈ 0.2606614021.
