Problem 11939
(American Mathematical Monthly, Vol.123, November 2016) Proposed by M. Omarjee (France).
Find
∞
X
k=1
1 +1
2 + · · · + 1
k − log(k) − γ − 1 2k + 1
12k2
Hereγ is Euler’s constant.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let Hk= 1 + 12+ · · · +1k, then
n
X
k=1
Hk =
n
X
k=1 k
X
j=1
1 j =
n
X
j=1
1 j
n
X
k=j
1 =
n
X
j=1
n + 1 − j
j = (n + 1)Hn− n.
Hence
n
X
k=1
Hk− log(k) − γ − 1 2k
= (n + 1)Hn− n − log(n!) − nγ − Hn
2
=
n +1
2
log(n) + γ + 1
2n+ O(1/n2)
− n − nγ
−
n log(n) − n +log(2π)
2 +log(n)
2 + O(1/n)
= 1 + γ
2 −
log(2π)
2 + O(1/n), where we used
Hn= log(n) + γ + 1
2n+ O(1/n2), and log(n!) = n log(n) − n +log(2π)
2 +log(n)
2 + O(1/n).
Moreover
∞
X
k=1
1 12k2 = 1
12· π2
6 = π2 72. Finally, we obtain
∞
X
k=1
Hk− log(k) − γ − 1 2k + 1
12k2
= 1 + γ
2 −
log(2π) 2 +π2
72.
