Showing that switching the order of variables gives same sum so as to simplify the required expression for exact sum If $S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$ , then if we want to show that $S(x,y) = S(y,x)$ so as to get the simplification by adding both to get the exact sum .               I tried to evaluate S(x,y)-S(y,x) , which gives $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)(5^y - 5^x)}{(5^{x+y}(5^x +5^y))}$ now how to do we show its value is zero ?

$S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$
By interchanging $x$ and $y$ we get
$S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^y(5^x +5^y))}$
$\implies 2S = \sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x +5^y)}\left(\frac{1}{5^x}+\frac{1}{5^y}\right)=\sum_{y=0}^{\infty}\sum_{x=0}^{\infty}\frac{(x+y +xy)}{(5^x 5^y)}$
$\implies  2S=\sum_{y=0}^{\infty}\sum_{x=0}^{\infty}\frac{(x+y +xy)}{(5^x 5^y)}$
Note that $\sum_{k=0}^{\infty} z^k=\frac{1}{1-z}$ and $\sum_{k=0}^{\infty}kz^k=\frac{z}{(1-z)^2},$ if $|z|<1$, then
$2S=2 \sum_{k=0}^{\infty} 5^{-k} \sum_{k=0}^{\infty}k 5^{-k}+\left(\sum_{k=0}^{\infty} k 5^{-k}\right)^2=\frac{25}{32}+\frac{25}{256}\implies S=\frac{223}{512}.$
