Compute explicitly $\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}(\frac{1}{2})^{x+y}$ It is possible to compute explicitly the following series?
$$\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}\Big(\frac{1}{2}\Big)^{x+y}$$
@Edit I tried to sum and subtract $2xy$ in the numerator.
In this way I get the following
\begin{align}
\sum_{x=1}^\infty\sum_{y=1}^\infty (x+y)\Big(\frac{1}{2}\Big)^{x+y}-2\sum_x x\Big(\frac{1}{2}\Big)^x\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
The first series should be easy to be computed. It remain the second one, in particular the following
\begin{align}
\sum_y\frac{y}{x+y}\Big(\frac{1}{2}\Big)^y=\sum_y\Big(\frac{1}{2}\Big)^y-x\sum_y\frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
Therefore the problem is reduced to computing:
\begin{align}
\sum_{y=1}^\infty \frac{1}{x+y}\Big(\frac{1}{2}\Big)^y.
\end{align}
But I don't know well how to do it.
 A: Little easier way is to do a substitution (since $n+m$ arrives many times). What I mean is, we can rewrite (where $x \in (0,1)$ let's say to not bother about changing the order of summation (but any $|x| < 1$ will be good due to absolute convergence of double series) $$ \sum_{m =1}^\infty \sum_{n=1}^\infty \frac{n^2 + m^2}{n+m} x^{n+m} = \sum_{m=1}^\infty \sum_{n=1}^\infty (n+m)x^{n+m} - 2\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{nm}{n+m}x^{n+m} $$
Now, let $k=n+m$. If $n,m$ can be any numbers between $1$ and $+\infty$, then $k$ is any number between $2$ and $+\infty$. If $k \ge 2$ is fixed, $n$ can be any number between $1$ and $k-1$, hence $$ \sum_{m=1}^\infty \sum_{n=1}^\infty (n+m)x^{n+m} = \sum_{k=2}^\infty \sum_{n=1}^{k-1} kx^k = \sum_{k=2}^\infty k(k-1)x^k $$
Similarly $$ \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{nm}{n+m}x^{n+m} = \sum_{k=2}^\infty \sum_{n=1}^{k-1} \frac{n(k-n)}{k}x^k = \sum_{k=2}^\infty \frac{x^k}{k} \sum_{n=1}^{k-1} n(k-n) = \frac{1}{6}\sum_{k=2}^\infty (k-1)(k+1)x^k $$
Both sums can be easily calculated via differentiating under sum sign, I will leave it to you
A: $$ \sum_{y=1}^\infty \frac{1}{x+y} \left(\frac{1}{2}\right)^y = \Phi(1/2,1,x) - 1/x$$
where $\Phi$ is the Lerch Phi function.
