# Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4,~a\not\in\mathbb Z$

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 expression1 in terms of the general parameter $$a\not\in\mathbb Z$$ ?

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

For somewhat similar expressions with 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 :

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

This question has already been asked on Mathematics Stack Exchange, 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$$)$$ hypergeometric functions, which, as already pointed out in the comment section to the original post, constitutes nothing more than a formal exercise in mathematical taxonomy.

• Formally, $S_4^-(a)$ is equal to $-a\,{}_5F_4(-a+1, -2a, -2a, -2a, -2a; -a, 1, 1, 1;1)$. This can be summed by Dougall's very well-poised $_5F_4$ summation formula (dlmf.nist.gov/16.4#E9) to give $$-\frac{a\Gamma(1+4a)}{\Gamma(1+2a)^3 \Gamma(1-2a)}.$$ Dec 9 '19 at 17:07
• Unfortunately, the command of Maple restart;s1:=sum((k + a)*binomial(2*a, -k)^4, k = -2*a .. infinity) assuming a>0, a::Not(integer); produces an incorrect result and the command of Mathematica Sum[(n + a)*Binomial[2*a, -n]^4, {n, -2*a, Infinity}, Assumptions -> a [Element] Reals && a [NotElement] Integers] performs the answer in terms of a hypergeometric function. Dec 9 '19 at 17:50
• In other words, $~S_3^-(a) = -\dfrac{\sin(2a\pi)}{2\pi}~$ and $~S_4^-(a) = -\displaystyle\frac{\sin(2a\pi)}{2\pi}\cdot{4a\choose 2a}~$ Dec 9 '19 at 17:51
• @IraGessel: Feel free to post an answer both here and on the original post. Dec 10 '19 at 3:39
• Just to be clear ... what is summation from $n=-2a$ when $a \notin \mathbb Z$? Dec 15 '19 at 13:10

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...
• The RHS satisfies $f(a+1) = \frac{f(a) (-4a)}{(2a+1)^2(2a+2)^2}$, check the LHS $g(a)$ satisfies the same, thus $h(a)=g(a)\Gamma(2a+1)^2\Gamma(-4a)-\frac1{4\cos(2\pi a)}$ is $1$-periodic, check it is entire function, show $\frac{h(a)}{a}$ vanishes as $a\to i\infty$ which implies $h(a)=O(a)$ thus $h(a)=h(0)+ah'(0)=h(0)$. @Lucian Dec 9 '19 at 22:01