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 ?

Your double series has non-negative terms and $\sum_{x\geq 0}\sum_{y\geq 0}=\sum_{y\geq 0}\sum_{x\geq 0}$ is granted by Fubini's theorem. Very crudely
$$ \sum_{x=0}^{M}\sum_{y=0}^{N}\frac{x+y+xy}{5^x(5^x+5^y)}\stackrel{\text{AM-GM}}{\leq}\sum_{x=0}^{M}\sum_{y=0}^{N}\frac{(1+x)(1+y)}{2\cdot 5^x 5^{\frac{x+y}{2}}}=\frac{1}{2}\sum_{x=0}^{M}\frac{x+1}{5^{3x/2}}\sum_{y=0}^{N}\frac{y+1}{5^{y/2}} $$
so
$$S=\sum_{y\geq 0}\sum_{x\geq 0}\frac{x+y+xy}{5^x(5^x+5^y)}=\frac{1}{2}\sum_{x,y\geq 0}\left(\frac{x+y+xy}{5^x(5^x+5^y)}+\frac{x+y+xy}{5^y(5^x+5^y)}\right)=\frac{1}{2}\sum_{x,y\geq 0}\frac{(x+1)(y+1)-1}{5^{x+y}}$$
or
$$ S = \frac{1}{2}\left(\sum_{x\geq 0}\frac{x+1}{5^x}\right)^2 - \frac{1}{2}\sum_{s\geq 0}\frac{s+1}{5^s}=\frac{1}{2}\cdot\frac{25}{16}\cdot\frac{9}{16}=\color{red}{\frac{225}{512}}. $$
