Problem 11921
(American Mathematical Monthly, Vol.123, June-July 2016) Proposed by C. I. V˘alean (Romania).
Prove
log2(2)
∞
X
k=1
Hk
(k + 1)2k+1 + log(2)
∞
X
k=1
Hk
(k + 1)22k +
∞
X
k=1
Hk
(k + 1)32k = ζ(4) + log4(2) 4 whereHk =Pk
j=11/j.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Since
[zk+1] log2(1 − z) = [zk+1]
−
∞
X
j=1
zj j
2
=
k
X
j=1
1
j(k + 1 − j)= 1 k + 1
k
X
j=1
1
j + 1
k + 1 − j
= 2Hk
k + 1, we have that for 0 < z < 1,P∞
k=1 Hkzk+1
k+1 = 12log2(1 − z). Then it suffices to show log(2)
∞
X
k=1
Hk
(k + 1)22k +
∞
X
k=1
Hk
(k + 1)32k =ζ(4) − log4(2)
4 . (1)
Now, for 0 < z ≤ 1, F (z) :=1 2
Z z 0
log2(1 − t)
t dt =
∞
X
k=1
Hkzk+1 (k + 1)2, and
Z z 0
F (t) t dt =
∞
X
k=1
Hkzk+1 (k + 1)3, and, by integrating by parts,
∞
X
k=1
Hkzk+1 (k + 1)3 =
Z z 0
F (t) d(log(t)) = [F (t) log(t)]z0+− Z z
0
F′(t) log(t) dt
= log(z)
∞
X
k=1
Hkzk+1 (k + 1)2 −1
2 Z z
0
log(t) log2(1 − t)
t dt.
Therefore, for 0 < z ≤ 1,
−2 log(z)
∞
X
k=1
Hkzk+1 (k + 1)2 + 2
∞
X
k=1
Hkzk+1 (k + 1)3 = −
Z z 0
log(t) log2(1 − t)
t dt (2)
and, by letting z =12, we have that (1) is equivalent to
− Z 12
0
log(t) log2(1 − t)
t dt = ζ(4) − log4(2)
4 . (3)
We notice that Z 12
0
log(t) log2(1 − t)
t dt = 1
2 Z 12
0
log2(1 − t) d(log2(t))
= 1
2log2(1 − t) log2(t)12
0++ Z 12
0
log(1 − t) log2(t) 1 − t dt
= log4(2)
2 +
Z 1
1 2
log(t) log2(1 − t)
t dt.
Hence, by using (2) for z = 1, we get
− Z 12
0
log(t) log2(1 − t)
t dt = −1
2 Z 12
0
log(t) log2(1 − t)
t dt −1
2
log4(2)
2 +
Z 1
1 2
log(t) log2(1 − t)
t dt
!
= −log4(2)
4 −1
2 Z 1
0
log(t) log2(1 − t)
t dt
= −log4(2)
4 +
∞
X
k=1
Hk
(k + 1)3 = ζ(4) − log4(2) 4 and the proof of (3) is complete because P∞
k=1 Hk
(k+1)3 = ζ(3, 1) = ζ(4)/4 which is a known result (see for example Special Values of Multidimensional Polylogarithms, Trans. Amer. Math. Soc. 353, 907-941, 2001 by Borwein, Bradley, Broadhurst, and Lisonek).
