Taking the logarithm the infinite product (which converges) becomes ∞ X n=2  −1 + (n2− 1) log  1 1 − 1/n2

Problem 11299

(American Mathematical Monthly, Vol.114, June-July 2007) Proposed by P. Refolio (Spain).

Show that

∞

Y

n=2

1 e

 n² n²− 1

ⁿ²⁻¹!

= e√e 2π .

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

Taking the logarithm the infinite product (which converges) becomes

∞

X

n=2



−1 + (n²− 1) log

 1

1 − 1/n²



=

∞

X

n=2

−1 + (n²− 1)

∞

X

k=1

1 kn^2k

!

=

∞

X

n=2

−1 +

∞

X

k=1

1 kn^2k−2−

∞

X

k=1

1 kn^2k

!

=

∞

X

n=2

∞

X

k=1

1 (k + 1)n^2k −

∞

X

k=1

1 kn^2k

!

=

∞

X

k=1

ζ(2k) − 1 k + 1 −

∞

X

k=1

ζ(2k) − 1 k Now the second series yields

∞

X

k=1

ζ(2k) − 1

k =

∞

X

n=2

∞

X

k=1

1 kn^2k =

∞

X

n=2

log

 n² n²− 1



= lim

N →∞log

N

Y

n=2

n² (n − 1)(n + 1)

!

= lim

N →∞log

 2N N + 1



= log 2.

So it suffices to prove that

∞

X

k=1

ζ(2k) − 1 k + 1 = 3

2− log π which is interesting by itself. Since the cotangent identity

cot x =

∞

X

n=0

1 x − nπ. holds in the interval 0 < x < π, then

x²cot x − x + x²

x − π − x²

x + π = 2

∞

X

n=2

x

x²− (nπ)² = −2

∞

X

n=2

x³/π² 1 − (x/(nπ))²

= −2x

∞

X

n=2

∞

X

k=1

 x nπ

^2k

= −2

∞

X

k=1

x^2k+1

π^2k (ζ(2k) − 1).

Therefore

∞

X

k=1

ζ(2k) − 1

k + 1 = 2

π² Z π

0

∞

X

k=1

x^2k+1

π^2k (ζ(2k) − 1) dx

= − 1 π²

Z π 0



x²cot x − x − x²

x − π− x² x + π

 dx

= 1

π² Z π

0



x + x² x + π



dx + 1 π² lim

t→π⁻

Z t 0

x² x − π dx −

Z t 0

x²cot x dx



= log 2 + 1 π² lim

t→π⁻

 x²

2 + πx + π²log(π − x)

t 0

− π²log(π − t) − π²log 2 + o(1)

!

= 3

2− log π

where we used the following asymptotic estimate as t → π⁻ Z t

0

x²cot x dx = Z t

0

x²d(log(sin x)) =x²log(sin x))^t

0− 2 Z t

0

x log(sin x) dx

= π²log((π − t)(1 + o(1))) − 2 Z t

0

x log(sin x) dx

= π²log(π − t) + π²log 2 + o(1)

and Z π

0

x log(sin x) dx = −π² 2 log 2.

The last integral can be evaluated in the following way Z π

0

x log(sin x) dx = Z π

0

(π − x) log(sin(π − x)) dx = π Z π

0

log(sin x) dx − Z π

0

x log(sin x) dx and therefore

Z π 0

x log(sin x) dx = π 2

Z π 0

log(sin x) dx = −π² 2 log 2 because

Z π 0

log(sin x) dx = 2 Z π/2

0

log(sin x) dx

= 2 Z π/2

0

log(sin(π/2 − x)) dx = 2 Z π/2

0

log(cos x) dx

= Z π/2

0

log(sin(2x)/2) dx = 1 2

Z π 0

log((sin x)/2) dx

= 1 2

Z π

0 log(sin x) dx −π 2log 2.