$\sum_{n=1}^\infty \sum_{m=1}^\infty\frac{1}{(m+n)^2}$ Does this double infinite sum $\sum_{n=1}^\infty \sum_{m=1}^\infty\frac{1}{(m+n)^2}$ converge?
I can not understand how Ido this
Please step by step
 A: In this sum, for each integer $t\ge2$, there are $t-1$ pairs $(m,n)$ with
$m+n=t$. So does
$$\sum_{t=2}^\infty\frac{t-1}{t^2}$$
converge?
A: Just for your curiosity since you received simple, clear and good answers.
Let us consider
$$S_p=\sum_{n=1}^p \sum_{m=1}^p\frac{1}{(m+n)^2}$$ we have
$$\sum_{m=1}^p\frac{1}{(m+n)^2}=\psi ^{(1)}(n+1)-\psi ^{(1)}(n+p+1)$$
$$S_p=2 \psi ^{(0)}(p+2)-\psi ^{(0)}(2 p+2)+2 (p+1) \psi ^{(1)}(p+2)-(2 p+1) \psi
   ^{(1)}(2 p+2)-\frac{\pi ^2}{6}+\gamma$$ Using the asymptotics of the polygamma functions, this would give
$$S_p=\left(\gamma +1-\log
   (2)-\frac{\pi ^2}{6}\right)+\log \left({p}\right)+\frac{3}{2 p}-\frac{35}{48 p^2}+O\left(\frac{1}{p^3}\right)$$
A: Just use comparison test. Firstly we know that inner sum converges (compare for example to $\sum \frac{1}{n^2}$), and so we can write
\begin{align}
\sum_{n=1}^k \sum_{m=1}^\infty\frac{1}{(m+n)^2} &\geq \sum_{n=1}^k \sum_{m=1}^\infty\frac{1}{(m+n)(m+n+1)}\\
&= \sum_{n=1}^k \sum_{m=1}^\infty\left(\frac{1}{m+n}-\frac{1}{m+n+1}\right)\\
&= \sum_{n=1}^k \frac{1}{n+1}\to \infty\text{ as }k\to \infty\\
\end{align}
The second equality follows by inner sum telescoping.
A: By comparision with an integral $$\frac 1{m^2}\ge\frac 1{m(m+1)}=\int_{m}^{m+1}\frac{dt}{t^2}$$
The double summation is in fact summing the remainder of the $\frac 1{n^2}$ series, but this remainder is only in $\frac 1n$, so the series diverge.
$\displaystyle\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty}\dfrac 1{(m+n)^2}=\sum\limits_{n=1}^{\infty}\sum\limits_{m=n+1}^{\infty}\dfrac 1{m^2}\ge\sum\limits_{n=1}^{\infty}\sum\limits_{m=n+1}^{\infty}\int_{m}^{m+1}\frac{dt}{t^2}\ge\sum\limits_{n=1}^{\infty}\int_{n+1}^{\infty}\frac{dt}{t^2}\ge\sum\limits_{n=1}^{\infty}\frac{1}{n+1}=\infty$
