
> 
> Do either $~S\_4^+(a)~=~\displaystyle\sum\_{n=0}^\infty(n+a){2a\choose n}^4~$ or $~S\_4^-(a)~=~\displaystyle\sum\_{n=-2a}^\infty(n+a){2a\choose-n}^4~$ possess a *meaningful* closed form expression**1** in terms of the general parameter $a\not\in\mathbb Z$ ?
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[Ramanujan](http://math.stackexchange.com/questions/2503975) provided the following [result](http://mathoverflow.net/questions/285746) : $~S\_4^-\Big(-\tfrac18\Big)~=~\dfrac1{\bigg[\Big(-\tfrac14\Big){\large!}\bigg]^2~\sqrt{8\pi}}~,~$ which would tentatively point to a possible closed form expression in terms of $~(2a)!~$ and/or $~(4a)!~$




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For somewhat similar expressions with lesser values of the exponent, we have [Dixon's identity](http://en.wikipedia.org/wiki/Dixon%27s_identity) :



> 
> $$\sum\_{n=0}^\infty(-1)^n{2a\choose n}^3 ~=~ \sum\_{n=-2a}^\infty(-1)^n{2a\choose-n}^3 ~=~ \cos(a\pi)~{3a\choose a,a},$$
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and [Vandermonde's identity](http://en.wikipedia.org/wiki/Vandermonde%27s_identity) :



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> $$\sum\_{n=0}^\infty(-1)^n{2a\choose n}^2 ~=~ \sum\_{n=-2a}^\infty(-1)^n{2a\choose-n}^2 ~=~ \cos(a\pi)~{2a\choose a},$$
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> $$\sum\_{n=0}^\infty{a\choose n}^2 ~=~ \sum\_{n=-a}^\infty{a\choose-n}^2 ~=~ {2a\choose a},$$
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as well as the [binomial theorem](http://en.wikipedia.org/wiki/Binomial_theorem) :



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> $$\sum\_{n=0}^\infty{a\choose n}^1x^n ~=~ (1+x)^a,\qquad\sum\_{n=-a}^\infty{a\choose-n}^1x^n ~=~ \Big(1+\tfrac1x\Big)^a.$$
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*This question has already been asked on [Mathematics Stack Exchange](http://math.stackexchange.com/questions/3412057), where, despite having been posted a considerable amount of time ago, and also put up for a bounty, it nonetheless still failed to receive a satisfying answer.*


**1** As opposed to, say, a mere rewriting in terms of $($[generalized](http://en.wikipedia.org/wiki/Generalized_hypergeometric_function)$)$ [hypergeometric functions](http://en.wikipedia.org/wiki/Hypergeometric_function), which, as already pointed out in the comment section to the original post, constitutes nothing more than a formal exercise in mathematical taxonomy.



Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4,~a\not\in\mathbb Z$
Some experimentation suggests that the second series is given by
$$
S\_4^-(a) = \sum\_{n=-2a}^\infty (n+a) \binom{2a}{-n}^4 = \frac{1}{4\cos(2\pi a) \,\Gamma(2a+1)^2\,\Gamma(-4a)},
$$
which agrees with Ramanujan's at $a = -1/8$, but I have no proof...


