Problem 11973
(American Mathematical Monthly, Vol.124, April 2017) Proposed by D. Orr (USA).
Prove
G = π 2
∞
X
n=0
ζ(2n) (2n + 1)4n
1 − 2
4n
, where G =P∞
n=0(−1)n/(2n + 1)2 (Catalan’s constant), ζ(s) = P∞
n=11/ns for s > 1, and ζ(0) =
−1/2.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. It is known that
(πx) cot(πx) = 1 + 2x2
∞
X
k=1
1
x2− k2 = 1 − 2
∞
X
k=1
(x/k)2 1 − (x/k)2
= 1 − 2
∞
X
k=1
∞
X
n=1
x2n
k2n = 1 − 2
∞
X
n=1
ζ(2n)x2n = −2
∞
X
n=0
ζ(2n)x2n. Hence
F (x) :=
Z πx
0
t cot(t) dt = π Z x
0
(πs) cot(πs) ds = −2π
∞
X
n=0
ζ(2n)x2n+1 2n + 1 . Moreover, for x ∈ [0, 1],
F (x) = −πxC1(2πx) +1
2S2(2πx) where
C1(θ) =
∞
X
k=1
cos(kθ)
k =
∞
X
k=1
eikθ+ e−ikθ 2k = −1
2ln(1 − eiθ) −1
2ln(1 − e−iθ)
= −1
2ln(2 − 2 cos(θ)) = − ln(|2 sin(θ/2)|), and
S2(θ) =
∞
X
k=1
sin(kθ) k2 .
In fact the above identity holds because l.h.s and r.h.s. agree at x = 0, and for x ∈ (0, 1), the derivative of the l.h.s., F′(x) = π2x cot(πx), coincides with the derivative of the r.h.s.,
d dx
−πxC1(2πx) +S2(2πx) 2
= −πC1(2πx) − 2π2xC1′(2πx) + πC2′(2πx)
= −πC1(2πx) − 2π2x(− cot(πx)/2) + πC1(2πx) = π2x cot(πx).
Hence π 2
∞
X
n=0
ζ(2n) (2n + 1)4n
1 − 2
4n
= π
∞
X
n=0
ζ(2n)(1/2)2n+1 (2n + 1) − 4π
∞
X
n=0
ζ(2n)(1/4)2n+1 (2n + 1)
= −F (1/2)
2 + 2F (1/4)
= πC1(π)
4 −S2(π)
4 −πC1(π/2)
2 + S2(π/2)
= π(− ln(2))
4 − 0 − π(− ln(√ 2))
2 + G = G.
