Problem 12102
(American Mathematical Monthly, Vol.126, March 2019) Proposed by O. Furdui and A. Sintamarian (Romania).
Prove
∞
X
n=1
Hn2 ζ(2) −
n
X
k=1
1 k2 − 1
n
!
= 2 − ζ(2) − 2ζ(3)
whereHn =Pn j=1
1
j is then-th harmonic number.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. For N ≥ 1
SN :=
N
X
n=1
Hn2 ζ(2) −
n
X
k=1
1 k2 −1
n
!
=
N
X
n=1
Hn2
∞
X
k=n+1
1 k2 −
N
X
n=1
Hn2
n
=
N
X
k=1
1 k2
k−1
X
n=1
Hn2−
N
X
n=1
Hn2
n + ζ(2) −
N
X
k=1
1 k2
!
·
N
X
n=1
Hn2.
It is easy to verify by induction thatPm−1
n=1 Hn2= mHm2 −(2m + 1)Hm+ 2m, therefore
N
X
k=1
1 k2
k−1
X
n=1
Hn2−
N
X
n=1
Hn2 n =
N
X
k=1
1
k2 kHk2−(2k + 1)Hk+ 2k
−
N
X
n=1
Hn2 n
= −2
N
X
k=1
Hk
k −
N
X
k=1
Hk
k2 + 2HN
= −HN2 −
N
X
k=1
1 k2 −
N
X
k=1
Hk
k2 + 2HN
= −HN2 + 2HN −ζ(2) − 2ζ(3) + o(1) where at the last step we used the known resultsP∞
n=1 1
k2 = ζ(2) andP∞
k=1 Hk
k2 = 2ζ(3).
On the other hand, since HN = ln(N ) + γ + o(1), andPN k=1
1
k2 = ζ(2) −N1 + O(1/N2), we have that ζ(2) −
N
X
k=1
1 k2
!
·
N
X
n=1
Hn2= 1
N + O(1/N2)
· (N + 1)HN2 −(2N + 1)HN + 2N
= HN2 −2HN + 2 + o(1).
Finally, as N → ∞,
SN = −HN2 + 2HN −ζ(2) − 2ζ(3) + HN2 −2HN + 2 + o(1) → 2 − ζ(2) − 2ζ(3).
