Problem 11715
(American Mathematical Monthly, Vol.120, June-July 2013) Proposed by M. Stofka (Slovakia).
Prove that
∞
X
k=0
1
(6k + 1)5 =1 2
25− 1
25 ·35− 1
35 ζ(5) + 11π5 8 · 35√
3
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
For 1 ≤ j ≤ 5, let
Sj=
∞
X
k=0
1 (6k + j)5. Then it is easy to see that
S3= 1 35
∞
X
k=0
1
(2k + 1)5 = 1
35 ζ(5) −
∞
X
k=1
1 (2k)5
!
=25− 1 65 ζ(5), S2+ S4= 1
25
∞
X
k=0
1 (3k + 1)5 +
∞
X
k=0
1 (3k + 2)5
!
= 1
25 ζ(5) −
∞
X
k=1
1 (3k)5
!
= 35− 1 65 ζ(5), S1+ S2+ S3+ S4+ S5= ζ(5) −
∞
X
k=1
1
(6k)5 = 65− 1 65 ζ(5).
Hence
S1+ S5=65− 35− 25+ 1
65 ζ(5) = 25− 1
25 ·35− 1 35 ζ(5).
By the the partial fraction expansion of the cotangent function, we know that for z ∈ C \ Z, π
tan(πz) =
∞
X
k=−∞
1 z − k.
Moreover, the following power series converges for all z such that |z| < 1, and since the sum of the terms with n > 1 converges absolutely, we may interchange the order of summation. So we have
F (z) :=
∞
X
n=1
∞
X
k=−∞
1 (6k + 1)n
! zn=
∞
X
k=−∞
∞
X
n=1
zn
(6k + 1)n = −z
∞
X
k=−∞
1
z − (6k + 1) = − πz
6 tan(π(z − 1)/6), which implies that
S1− S5=
∞
X
k=−∞
1
(6k + 1)5 = 1 5!
d5F
dz5(0) = 11π5 8 · 35√
3. Finally, we obtain
S1=(S1+ S5) + (S1− S5)
2 =1
2
25− 1
25 ·35− 1
35 ζ(5) + 11π5 8 · 35√
3
, S5=(S1+ S5) − (S1− S5)
2 =1
2
25− 1
25 ·35− 1
35 ζ(5) − 11π5 8 · 35√
3
.
