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!!!