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Prove that $\sum\limits_{m,n=1}^{\infty} \dfrac{1}{m^pn^q}$ converges if $p>1$ and $q>1$.

I am not sure how to really approach this (I don't really know anything about double series) but this is what I tried,

Let $b_m = \sum\limits_{n=1}^{\infty} \dfrac{1}{m^pn^q}$ so

$b_1 = \sum\limits_{n=1}^{\infty} \dfrac{1}{n^q}$ convergent p-series

$b_2 = \dfrac{1}{2^p}\sum\limits_{n=1}^{\infty} \dfrac{1}{n^q}$ convergent p-series

$b_3 = \dfrac{1}{3^p}\sum\limits_{n=1}^{\infty} \dfrac{1}{n^q}$ convergent p-series

Let $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^q}=L$ since it is convergent set it equal to its limit. Now look at

$\sum\limits_{m=1}^{\infty} b_m = L + \dfrac L{2^p} + \dfrac L{3^p} + \dfrac L{4^p} + \cdots = L \left( 1 + \dfrac 1{2^p} + \dfrac 1{3^p} + \dfrac 1{4^p} + \right) = L\sum\limits_{m=1}^{\infty} \dfrac 1{m^p}$

which is another convergent p-series and therefore the double sum converges. Is this right or am I going about this completely wrong? If I am wrong can someone please explain it to me. Thank you!!!

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It looks right for the most part, but you have to argue (it shouldn't be too hard) that the limit doesn't change when you take the limit of $n$ first as opposed to $n$. –  Clayton Dec 14 '12 at 2:24
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1 Answer

In this case, it's easy because the double series factors as the product of two ordinary series, and because these are series with positive terms: $$ \sum_{m=1}^M \sum_{n=1}^N \frac{1}{m^p n^q} = \left( \sum_{m=1}^M \frac{1}{m^p} \right) \left( \sum_{n=1}^N \frac{1}{n^q} \right)$$ so it converges if both of those series on the right converge, and diverges to $+\infty$ if either of the series on the right diverges to $+\infty$.

In cases involving both positive and negative terms, things might be trickier. Actually it's not clear in what order you want to take the terms in a double series, so conditional convergence is rather ambiguous.

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What about if its in the form $\sum\limits_{m,n=1}^{\infty} \dfrac{1}{(m^2 +n^2)^p}$ converges if $p>1$? –  user52709 Dec 14 '12 at 2:29
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Compare that one to a double integral, and change to polar coordinates. –  Robert Israel Dec 14 '12 at 2:33
    
I am not sure I understand what you mean about comparing it. –  user52709 Dec 14 '12 at 2:39
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$1/(m^2 + n^2)^p$ is greater than the integral of $1/(x^2 + y^2)^p$ over $[m,m+1] \times [n,n+1]$, and less than the integral over $[m-1,m] \times [n-1,n]$. –  Robert Israel Dec 14 '12 at 2:45
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