Problem 12222
(American Mathematical Monthly, Vol.127, December 2020) Proposed by R. Tauraso (Italy).
Prove
∞
X
k=1
(−1)k k2
∞
X
n=k
1
n2n = −13ζ(3) 24 .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Since Z 1
0
tklog(t) dt = tk+1 k + 1log(t)
1
0
− Z 1
0
tk+1−1
k + 1 dt = − 1 (k + 1)2, it follows that the given double series has an integral representation
I :=
Z 1
0
log(t + 2) log(t) t + 1 dt =
Z 1
0
(log(2) + log(1 +2t)) log(t)
t + 1 dt
= log(2) Z 1
0
log(t) t + 1 dt +
∞
X
n=1
(−1)n−1 n2n
Z 1 0
tnlog(t) t + 1 dt
= log(2) Z 1
0
log(t)
∞
X
k=0
(−1)ktkdt +
∞
X
n=1
(−1)n−1 n2n
Z 1
0
tnlog(t)
∞
X
k=0
(−1)ktkdt
= − log(2)
∞
X
k=0
(−1)k (k + 1)2 −
∞
X
n=1
(−1)n−1 n2n
∞
X
k=0
(−1)k (n + k + 1)2
= −log(2)ζ(2)
2 +
∞
X
n=1
1 n2n
ζ(2) 2 +
n
X
k=1
(−1)k k2
!
= −log(2)ζ(2) 2 +ζ(2)
2
∞
X
n=1
1 n2n +
∞
X
n=1
1 n2n
n
X
k=1
(−1)k k2
= −log(2)ζ(2)
2 +log(2)ζ(2)
2 +
∞
X
n=1
1 n2n
n
X
k=1
(−1)k k2
=
∞
X
k=1
(−1)k k2
∞
X
n=k
1 n2n.
Now we evaluate the integral I. We introduce two ausiliary functions F− and F+: F−(x) :=
Z x 0
log2(t)
1 − t dt = − log2(x) log(1 − x) + 2 Z x
0
log(t) log(1 − t)
t dt
= − log2(x) log(1 − x) − 2 Z x
0
log(t)
∞
X
k=1
tk−1 k dt
= − log2(x) log(1 − x) − 2 log(x)Li2(x) + 2 Z x
0
∞
X
k=1
tk−1 k2 dt
= − log2(x) log(1 − x) − 2 log(x)Li2(x) + 2Li3(x) and, in a similar way,
F+(x) :=
Z x 0
log2(t)
1 + t dt = log2(x) log(1 + x) + 2 log(x)Li2(−x) − 2Li3(−x) . Hence
I = Z 1
0
log(t + 2) log(t) t + 1 dt =1
2 Z 1
0
log2(t)
t + 1 +log2(t + 2) t + 1 −
log2
t t+2
t + 1
dt.
The first term is
Z 1 0
log2(t)
t + 1 dt = F+(1) = −2Li3(−1) =3ζ(3) 2 . Moreover, by letting s = 1/(t + 2) we get
Z 1 0
log2(t + 2) t + 1 dt = −
Z 1/3 1/2
log2(s) s(1 − s)ds =
Z 1/2 1/3
log2(s) 1 − s ds +
Z 1/2 1/3
log2(s) s ds
= F−(1/2) − F−(1/3) +log3(3)
3 −
log3(2) 3 and, by letting s = t/(t + 2) we find
Z 1 0
log2
t t+2
t + 1 dt = 2
Z 1/3 0
log2(s) 1 − s2 ds =
Z 1/3 0
log2(s) 1 − s ds +
Z 1/3 0
log2(s) 1 + s ds
= F−(1/3) + F+(1/3).
Therefore I = 3ζ(3)
4 +log3(3)
6 −
log3(2)
6 +F−(1/2)
2 − F−(1/3) −F+(1/3) 2
=13ζ(3)
8 −
log3(3)
3 − log(3)
2Li2
1 3
− Li2
− 1 3
−
2Li3
1 3
− Li3
− 1 3
. Since it is known that (see Lewin’s book Polylogarithms and Associated Functions)
2Li2
1 3
− Li2
− 1 3
= π2 6 −
log2(3) 2 and
2Li3
1 3
− Li3
− 1 3
= log3(3)
6 −
log(3)π2
6 +13ζ(3) 6 we finally find
I = −13ζ(3) 24 .
