# Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4$

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 ?

Ramanujan provided the following result : $$~S_4^-\Big(-\tfrac18\Big)~=~\dfrac1{\bigg[\Big(-\tfrac14\Big){\large!}\bigg]^2~\sqrt{8\pi}}~,~$$ which would point to a possible closed form expression in terms of $$~(2a)!~$$ and/or $$~(4a)!~$$

For lesser values of the exponent, we have Dixon's identity :

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

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

$$\sum_{n=0}^\infty{a\choose n}^2 ~=~ \sum_{n=-a}^\infty{a\choose-n}^2 ~=~ {2a\choose a},$$

as well as the binomial theorem :

$$\sum_{n=0}^\infty{a\choose n}^1x^n ~=~ (1+x)^a,\qquad\sum_{n=-a}^\infty{a\choose-n}^1x^n ~=~ \Big(1+\tfrac1x\Big)^a.$$

• First of all, good to see you back here ! I suppose that you do not want hypergeometric functions. Right ? Cheers – Claude Leibovici Oct 28 '19 at 7:28
• @ClaudeLeibovici: Same here. :-) And no, since the latter are simply an exercise in mathematical taxonomy, so to say. – Lucian Oct 28 '19 at 7:32
• I was sure of that. – Claude Leibovici Oct 28 '19 at 7:35
• Now posted to MO, mathoverflow.net/questions/347906/… – Gerry Myerson Dec 9 '19 at 0:49

When $$2a$$ is a non-negative integer: $$S=\sum_{n=-2a}^{\infty} (n+a) {2a \choose -n}^4= \sum_{n=0}^{2a}(-n+a) {2a \choose n}^4=\sum_{n=0}^{2a}~(-(2a-n)+a)~{2a\choose n}^4$$ $$=\sum_{n=0}^{2a} (n-a) {2a \choose n}^4~ =-S.$$ So $$S=-S \implies S=0$$.Here we have used : $${n \choose-k}= 0~$$ if $$k\in I^+$$, $$\sum_{k=0}^{n} f(k) =\sum_{k=0}^n f(n-k)$$, and $${n \choose k}={n \choose n-k}$$
$$S_4^-(a)~=~S_3^-(a)\cdot{4a\choose2a}~=~-\frac{\sin(2a\pi)}{2\pi}\cdot{4a\choose2a}$$