Problem 11494
(American Mathematical Monthly, Vol.117, March 2010) Proposed by O. Furdui (Romania).
LetA denote the Glaisher-Kinkelin constant, given by
A = lim
n→∞n−n2/2−n/2−1/12en2/4
n
Y
k=1
kk= 1.2824 . . .
Prove that
∞
Y
n=1
n!
√2πn(n/e)n
(−1)n−1
= A3 27/12π1/4.
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 the infinite product can be written as
∞
Y
n=1
(2n − 1)!
p2π(2n− 1)((2n − 1)/e)2n−1
!
·
(2n)!
√2π2n(2n/e)2n
−1
=
∞
Y
n=1
1 e
2n 2n − 1
2n−1/2
. Hence it suffices to prove that
SN :=
N
X
n=1
−1 + (2n − 1
2)(log(2n) − log(2n − 1))
= 3 log(A) − 7
12log(2) − 1
4log(π) + o(1).
By expanding the partial sum we obtain SN = −N +
N
X
n=1
(2n) log(2n) −1 2
N
X
n=1
log(2n) −
N
X
n=1
(2n − 1) log(2n − 1) −1 2
N
X
n=1
log(2n − 1)
= −N + 2
N
X
n=1
(2n) log(2n) −
2N
X
n=1
n log(n) +1 2
2N
X
n=1
log(n)
= −N + 4 log(2)
N
X
n=1
n + 4
N
X
n=1
n log(n) −
2N
X
n=1
n log(n) −1 2
2N
X
n=1
log(n).
By Stirling’s approximation
2N
X
n=1
log(n) = 1
2log(2π) +1
2log(2N ) + (2N ) log(2N ) − 2N + o(1)
and by the constant A’s definition we have that
N
X
n=1
n log(n) = log(A) + N2 2 +N
2 + 1 12
log(N ) −N2 4 + o(1)
and 2N
X
n=1
n log(n) = log(A) +
2N2+ N + 1 12
log(2N ) − N2+ o(1).
Finally,
SN = −N + 2 log(2)N(N + 1) + 4 log(A) +
2N2+ 2N +1 3
log(N ) − N2
− log(A) −
2N2+ N + 1 12
log(2N ) + N2− 1
4log(2π)
− 1
4log(2N ) − N log(2N) + N + o(1) = 3 log(A) − 7
12log(2) −1
4log(π) + o(1).
