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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 − e) −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.



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