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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!

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2 Answers 2

up vote 2 down vote accepted

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.

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Thanks for the clean approach! –  Vladimir Feb 19 '13 at 3:10

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 $Q$.

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