2 N −1 X n=1 22n−1 2n + 1  (n − 1)! (2n − 1

Problem 11952

(American Mathematical Monthly, Vol.124, January 2017) Proposed by Z. K. Silagadze (Russia).

Prove that

∞

X

n=1

2²ⁿ⁻¹ 2n + 1

 (n − 1)!

(2n − 1)!!

²

= π − 2.

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

Solution. We first show by induction that for N > 0,

SN := 2

N −1

X

n=1

2²ⁿ⁻¹ 2n + 1

 (n − 1)!

(2n − 1)!!

²

=

N −1

X

n=1

4²ⁿ n²(2n + 1) ²ⁿn

² = (2N − 1)4^2N N^{2 2N}_N
² − 4.

The identity is true for N = 1. For N > 1 we have to verify that 4^2N

N²(2N + 1) ^2NN

² = SN +1− S^N =





(2N + 1)4^{2N +2} (N + 1)^{2 2N +2}_{N +1}
² − 4



− (2N − 1)4^2N N^{2 2N}_N
² − 4

! ,

that is

1

N²(2N + 1)= 4

(2N + 1)−(2N − 1) N² , which holds. Hence

∞

X

n=1

2²ⁿ⁻¹ 2n + 1

 (n − 1)!

(2n − 1)!!

²

= 1 2 lim

N →+∞SN = 1 2 lim

N →+∞

(2N − 1)4^2N

N^{2 2N}_N
² − 2 = π − 2 because by Stirling approximation

N →+∞lim

√N 4^N

2N N



= 1

√π.