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

[Olivier Gerard]

The good news is that identity is true.
Unfortunately, this equality is known
and can be proved by known methods.
The way you wrote it can just make things
a little less tractable.

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])


Sum[ 1/((2 n + 1)!/(n!)^2), {n, 0, Infinity}]

(2 \[Pi])/(3 Sqrt[3])

The sum gives your original value.


On Sat, Aug 18, 2012 at 10:59 PM, Alexander P-sky <apovolot at gmail.com> wrote:
> Hi,
>
> It appears that I have stumbled upon new (?)  identity (?) involving
> Pi and sqrt(3)
>
> sum(1/(n!/floor(n/2)!^2),n=0...infinity) = 4/3 + 8*Pi/(9*sqrt(3))
>
> Both left and right sides give something like
> 2.94559943487486031163918067345969398425250333163799162272878660923388727211231456327477776729960903898331
>
> I am using at the moment WolframAlpha Pro (just for two weeks of free
> evaluation)
> It (WolframAlpha Pro) times out on me in proving or disproving
> whether this identity is true or false.
>
> Could some one check it out, please ?
>
> Thanks,
> Best Regards,
> Alexander R. Povolotsky