Let τ = i/√ 3 and q = exp(2πiτ ) then ∞ Y n=1  1 + 2e−nπ √3cosh nπ √3

Problem 11677

(American Mathematical Monthly, Vol.119, December 2012) Proposed by Albert Stadler (Switzerland).

Evaluate

∞

Y

n=1



1 + 2e^−nπ

√3cosh nπ

√3



.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let τ = i/√

3 and q = exp(2πiτ ) then

∞

Y

n=1



1 + 2e^−nπ

√3cosh nπ

√3



=

∞

Y

n=1

1 + qⁿ+ q²ⁿ

= Q∞

n=1(1 − q³ⁿ) Q∞

n=1(1 − qⁿ)

= exp(−πiτ /6)η(3τ ) η(τ )

= exp(−πiτ /6)η(−1/τ ) η(τ )

= exp(−πiτ /6)√

−iτ

=exp(π√ 3/18) 3^1/4 where we used the following identity which holds for Im(τ ) > 0

η(−1/τ ) =√

−iτ η(τ ) with

η(τ ) = exp(πiτ /12)

∞

Y

n=1

(1 − exp(2πinτ )),

the Dedekind eta function.