Convergence of a particular double series For the double series
$$ \sum_{m,n=1}^{\infty} \frac{1}{(m+n)^p} , $$
I was wondering when it converges. I want to use double integrals to estimate it, but I don't know how to write the process accurately... Could you show me a detailed computation? Thanks!
 A: While we wait for someone to do it using double integrals, here's another way:
Given a positive integer $k$, the number of pairs $m,n$ with $m\ge1$, $n\ge1$, and $m+n=k$ is $k-1$. So your sum is $\sum_{k=2}^{\infty}(k-1)/k^p$, and the usual single-series methods apply. 
A: If you really want a double integral approach (though I think Gerry Myerson's solution is much better), here goes:
Let $$f(x,y) = \frac{1}{(x+y)^p}$$
Note that
$$0 \le f(m,n) \le \iint_{Q_{m,n}} f(x,y)\,dx\,dy$$
where $Q_{m,n} = \{ (x,y) : m-1 \le x \le m, n-1 \le y \le n \}$, since $f$ is decreasing in $x$ and $y$ separately. (We can assume that $p > 0$, since otherwise the double series is obviously divergent.)
There is a technical problem here though. We don't want to use the inequality above when $m = 0$ or $n= 0$ since $f$ is unbounded on $Q_{0,n}$ and $Q_{m,0}$. But since all terms of your series are positive, we may throw away
$$\sum_{n=1}^\infty \frac{1}{m^p} \qquad\text{and}\qquad \sum_{m=2}^\infty \frac{1}{n^p}.$$
These converge if and only if $p > 1$, so this is certainly necessary for the double series to converge. (And we may estimate these one-variable series using one-variable techniques. I will leave the details to you.)
For the remainder, summing up, we see that
\begin{align}
\sum_{m,n=2}^\infty \frac{1}{(m+n)^p} &\le
\iint_Q \frac{1}{(x+y)^p}\,dx\,dy \\
&= \int_1^{\infty} \left( \int_1^\infty \frac{1}{(x+y)^p}\,dx \right)\,dy\\
&= \frac{1}{p-1} \int_1^{\infty} \frac{1}{(y+1)^{p-1}}\,dy \\
&= \frac{4}{2^p(p-1)(p-2)}
\end{align}
with convergence exactly when $p > 2$. (When $p \le 2$, you can turn the estimates the other way around to show that your series diverges.)
Here, $Q$ is the union of all $Q_{m,n}$ for $m,n \ge 2$, i.e. $Q = \{ (x,y) : x \ge 1, y \ge 1 \}$, and all the computations are justified since $f \ge 0$ on $Q$.
