# Difficult Sum: Any Tips?

I'm trying to integrate a very difficult expression, and I've arrived at a particular step involving this sum, which I want to turn into a closed expression so that I can raise it to a power, re-sum over one of the variables and then integrate. Here's the sum:

$$\sum_{n=0}^{\infty}\sum_{m=0}^{n}\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\left(\left(\frac{(-1)^{n}(2n)!(\beta^{n})(r^{2n-2m})}{n!(2m)!(2n-2m)!}\right)\left(\frac{(\frac{1}{2})_{i+j}(\frac{1}{2}+i-j)_{i}(-i)_{jj}}{\left(\frac{3}{2}\right)_{i+j}i!j!}\right)\left(\frac{g^{2(i+n)+1}}{r^{2i+1}}\right)\right)$$

where $\left(a\right)_{n}$ is the Pochhammer symbol (essentially I have written out explicitly an F1 Appell Hypergeometric Series into the middle and final brackets); as such one can rewrite the expression as:

$$\sum_{n=0}^{\infty}\sum_{m=0}^{n}\left(\left(\frac{(-1)^{n}(2n)!(\beta^{n})(r^{2n-2m})}{n!(2m)!(2n-2m)!}\right)AppellF1[\frac{1}{2},\frac{1}{2}+m-n,-m,\frac{3}{2},(\frac{g}{r})^{2},1]\left(\frac{g^{2n+1}}{r}\right)\right)$$

If you saw this and wanted to try and simplify it as much as possible so that raising (the whole expression) to a particular power would be possible (avoiding the need to use the infinite multinomial theorem), how would you go forward? Does it look too complex to have a closed form expression?

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p.s. The sum definitely converges. – Alexander Giles Sep 4 '13 at 20:44
g and beta are constants – Alexander Giles Sep 5 '13 at 0:09
Does it look too complex to have a closed form expression? No, it's too easy :p. This expression needs a miracle. – whatever Aug 6 '14 at 20:26
This is easy. All you have to do is add some gamma-zero of (M) structure and just run through your calculations. – David Simmons Apr 14 at 15:10
Of course! Ty David Simmons! – Alexander Giles Apr 14 at 16:09