I have tried well over 30 cumulative hours trying to evaluate this double series:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{k+j+m+1}},[/tex]

for some integer [tex]m>0[/tex]. I have simplified (or maybe complicated) the sum to

[tex]S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}\frac{m!}{k!(m-k)!}\frac{(k+m-1-j)!}{(k+m)!}\frac{(k+m-1/2)!}{(k+m-1/2-j)!}\frac{1}{2^{k-j+m+1}},[/tex]

which may be easier to work with (hopefully it is correct).

I conjecture that [tex]S=0[/tex] from calculating [tex]S[/tex] with 50 consecutive values of [tex]m[/tex]. And assuming Mathematica knows what it's doing, [tex]m[/tex] has to equal zero from earlier results; results preceding the derivation of this double sum.

I am wondering if anyone knows an easy way to evaluate this sum. :)

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# Difficult Double Series

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