[seqfan] Re: It appears that I have stumbled upon new (?) identity (?) involving Pi and sqrt(3)

zak seidov zakseidov at yahoo.com
Sun Aug 19 11:47:17 CEST 2012





----- Original Message -----
> From: Olivier Gerard <olivier.gerard at gmail.com>

<...> 
> By the way, if you have access to Wolfram Alpha, you will see that it can derive
> it if you split it on even and odd n:
> 
> Sum[ 1/((2 n)!/n!^2), {n, 0, Infinity}]
> 
> 2/27 (18 + Sqrt[3] \[Pi])



Also, see Comment to A000984Central binomial coefficients: C(2*n,n) = (2*n)!/(n!)^2 :um(n>=1, 1/a(n))=(2*Pi*sqrt(3)+9)/27 - Benoit Cloitre, May 01 2002 

Zak Seidov



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