Problem 12206
(American Mathematical Monthly, Vol.127, October 2020) Proposed by S. Stewart (Australia).
Prove
∞
X
n=1
H2n
n2 = 3 4ζ(3) whereHn is then-th skew-harmonic numberPn
k=1(−1)k−1/k.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let Hn(x) =Pn
k=1xk/k. Then Hn(1) = Hn and Hn(−1) = −Hn. Moreover, for 0 < |x| ≤ 1,
∞
X
n=1
Hnxn n2 =
∞
X
n=1
xn n2
∞
X
k=1
1 k − 1
n + k
=
∞
X
n=1
∞
X
k=1
xn nk(n + k)
=
∞
X
k=1
∞
X
n=k+1
xn−k (n − k)kn =
∞
X
n=2 n−1
X
k=1
xn−k (n − k)kn =
∞
X
n=2
1 n2
n−1
X
k=1
xn−k
k + xn−k n − k
=
∞
X
n=1
Hn−1(1/x)xn
n2 +
∞
X
n=1
Hn−1(x) n2 . Therefore, for x = 1 we get
∞
X
n=1
Hn n2 = 2
∞
X
n=1
Hn−1
n2 = 2
∞
X
n=1
Hn
n2 − 2ζ(3) =⇒
∞
X
n=1
Hn
n2 = 2 ζ(3), and for x = −1 we find
∞
X
n=1
(−1)nHn n2 = −
∞
X
n=1
(−1)nHn−1
n2 −
∞
X
n=1
Hn−1
n2 = −
∞
X
n=1
((−1)n+ 1)Hn−1
n2
= −2
∞
X
n=1
H2n−1
(2n)2 = −1 2
∞
X
n=1
H2n
n2 −1
4ζ(3). (1)
On the other hand,
∞
X
n=1
H2n
n2 =
∞
X
n=1
1 n2
n
X
k=1
1 2k − 1−
n
X
k=1
1 2k
!
=
∞
X
n=1
1 n2
H2n−Hn 2
−Hn 2
= 4
∞
X
n=1
H2n
(2n)2 −
∞
X
n=1
Hn n2 = 2
∞
X
n=1
((−1)n+ 1)Hn
n2 −
∞
X
n=1
Hn n2
= 2
∞
X
n=1
(−1)nHn n2 +
∞
X
n=1
Hn n2 = 2
∞
X
n=1
(−1)nHn
n2 + 2 ζ(3). (2)
By solving the linear system given by (1) and (2) we easily obtain
∞
X
n=1
H2n
n2 = 3
4ζ(3), and
∞
X
n=1
(−1)nHn n2 = −5
8ζ(3).
Remark: all the above infinite series are absolutely convergent and therefore the terms can be
rearranged in a different order.
