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

[israel at math.ubc.ca]

In fact both of these series can be obtained from generating functions

sum_{n=0}^infty (n!^2/(2 n)!) z^n 
    = 4 sqrt(z/(4-z)^3) arcsin(sqrt(z)/2) + 4/(4-z)

sum_{n=0}^infty (n!^2/(2 n + 1)!) z^n 
    = (4/sqrt(4z - z^2)) arcsin(sqrt(z)/2)

Robert Israel
University of British Columbia

On Aug 18 2012, Olivier Gerard wrote:

>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
>
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