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> 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 in terms of the general parameter *a* ?
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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 point to a possible closed form expression in terms of $~(2a)!~$ and/or $~(4a)!~$

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

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> $$\sum\_{n=0}^\infty(-1)^n{2a\choose n}^3 ~=~ \sum\_{n=-2a}^\infty(-1)^n{2a\choose-n}^3 ~=~ \cos(a\pi)~{3a\choose a,a},$$
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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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