·(N − 1)N2−1· (N + 1)N2−1 N2(N2−1

Problem 11333

(American Mathematical Monthly, Vol.114, December 2007) Proposed by P. F. Refolio (Spain).

Show that

∞

Y

n=2

 n²− 1 n²

^2(n2−1) n + 1 n − 1

ⁿ!

= π.

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

First we consider the finite telescopic product

N

Y

n=2

 n + 1 n − 1

ⁿ

= 3² 1² ·4³

2³· 5⁴

3⁴· · · N^{N −1}

(N − 2)^{N −1} ·(N + 1)^N (N − 1)^N

= N^{N −1}(N + 1)^N

2((N − 1)!)² = N^{2N +1} 2(N !)²

 1 + 1

N

^N

= e^{2N +1}

4π (1 + o(1)) where in the last step we used

N ! =√

2πN (N/e)^N(1 + o(1)) and (1 + 1/N )^N = e(1 + o(1)).

Now we consider another finite telescopic product

N

Y

n=2

 n²− 1 n²

ⁿ²⁻1

=

N

Y

n=2

(n − 1)ⁿ²⁻¹(n + 1)ⁿ²⁻¹ n^2(n2−1)

= 1³· 3³

2⁶ ·2⁸· 4⁸

3¹⁶ ·3¹⁵· 5¹⁵

4³⁰ · · ·(N − 1)^N2−1· (N + 1)^N2−1 N^2(N2−1)

= ((N − 1)!)²· (N + 1)^N2−1

N^{N2+2N −2} = ((N )!)² N^{2N +1} ·

 1 + 1

N

^N2−1

= 2π

e^2N · e^{N −12}(1 + o(1)) = 2π

e^N+12(1 + o(1)) where in the last step we used the Stirling approximation and

 1 + 1

N

^N2−1

= exp



N²− 1
 1 N − 1

2N² + o(1/N²)



= exp

 N −1

2 + o(1)



= e^{N −12}(1+o(1)).

Finally

N

Y

n=2

 n²− 1 n²

^2(n2−1) n + 1 n − 1

ⁿ!

=

N

Y

n=2

 n²− 1 n²

ⁿ²⁻¹!²

·

N

Y

n=2

 n + 1 n − 1

ⁿ

= 4π²

e^{2N +1}(1 + o(1)) ·e^{2N +1}

4π (1 + o(1)) → π.