TITLE: Difficult Sum: Any Tips?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: Hint:
For $\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\left(\dfrac{(-1)^n(2n)!\beta^nr^{2n-2m}}{n!(2m)!(2n-2m)!}F_1\left(\dfrac{1}{2},\dfrac{1}{2}+m-n,-m,\dfrac{3}{2};\dfrac{g^2}{r^2},1\right)\dfrac{g^{2n+1}}{r}\right)$ ,
$\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\left(\dfrac{(-1)^n(2n)!\beta^nr^{2n-2m}}{n!(2m)!(2n-2m)!}F_1\left(\dfrac{1}{2},\dfrac{1}{2}+m-n,-m,\dfrac{3}{2};\dfrac{g^2}{r^2},1\right)\dfrac{g^{2n+1}}{r}\right)$
$=\sum\limits_{m=0}^\infty\sum\limits_{n=m}^\infty\left(\dfrac{(-1)^n(2n)!\beta^ng^{2n+1}r^{2n-2m-1}}{n!(2m)!(2n-2m)!}{_2}F_1\left(\dfrac{1}{2},-m;\dfrac{3}{2};1\right){_2}F_1\left(\dfrac{1}{2},\dfrac{1}{2}+m-n;m+\dfrac{3}{2};\dfrac{g^2}{r^2}\right)\right)$ (according to http://functions.wolfram.com/HypergeometricFunctions/AppellF1/03/04)
$=\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\left(\dfrac{(-1)^{m+n}(2m+2n)!\beta^{m+n}g^{2m+2n+1}r^{2n-1}\Gamma\left(\dfrac{3}{2}\right)\Gamma(m+1)}{(m+n)!(2m)!(2n)!\Gamma\left(m+\dfrac{3}{2}\right)}{_2}F_1\left(\dfrac{1}{2},\dfrac{1}{2}-n;m+\dfrac{3}{2};\dfrac{g^2}{r^2}\right)\right)$ (according to https://en.wikipedia.org/wiki/Hypergeometric_function#Special_values_at_z_=_1)
$=\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\left(\dfrac{(-1)^{m+n}\left(\dfrac{3}{2}\right)_{m+n}\beta^{m+n}g^{2m+2n+1}r^{2n}}{\left(\dfrac{3}{2}\right)_m\left(\dfrac{3}{2}\right)_mn!\left(\dfrac{3}{2}\right)_n\sqrt{r^2-g^2}}{_2}F_1\left(\dfrac{1}{2},m+n+1;m+\dfrac{3}{2};\dfrac{g^2}{g^2-r^2}\right)\right)$ (according to http://mathworld.wolfram.com/PochhammerSymbol.html)