Why does this series converge? My question is:


  Why does the series $$ \sum_{j,k=1}^\infty \frac{1}{j^4+k^4} $$ converge?


I tested the convergence with Mathematica and Octave, but I can't find an analytical proof. In fact, numerical computations suggest that the value of the series is $&lt;1$.



One obvious thing to do would be to use the generalized harmonic series to see that \begin{align} \sum_{j,k=1}^\infty \frac{1}{j^4+k^4} &amp;= \sum_{k=1}^\infty \frac{1}{2k^4} + \sum_{j,k=1; j\neq k} \frac{1}{j^4+k^4} \\ &amp;= \frac{\pi^4}{180} + \sum_{j=1}^\infty \sum_{k=1}^{j-1} \frac{1}{j^4+k^4} + \sum_{j=1}^\infty \sum_{k=j+1}^{\infty} \frac{1}{j^4+k^4}\\ &amp;\leq \frac{\pi^4}{180} + \sum_{j=1}^\infty \sum_{k=1}^{j-1} \frac{1}{(j-k)^4} + \sum_{j=1}^\infty \sum_{k=j+1}^{\infty} \frac{1}{(j-k)^4} \end{align} but unfortunately the last two (double-)series do not converge.



The problem arises when one tries to estimate the Hilbert-Schmidt norm of the Laplacian in $H^2(\mathbb{T}_\pi^2)$.
You can use the inequality $2xy\le x^2+y^2$ (seen by expanding $(x-y)^2\ge0$) to get $2j^2k^2\le j^4+k^4$:
$$\sum_{j,k=1}^\infty\frac1{j^4+k^4}\le\frac12\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{j^2}\frac1{k^2}=\frac12\Bigl(\sum_{j=1}^\infty\frac1{j^2}\Bigr)^2&lt;\infty.$$
Straightforward application of comparison test: 
$$
\sum_{j,k=1}^\infty \frac{1}{j^4+k^4} &lt; \sum_{j=1}^\infty \frac{1}{j^4} = \frac{\pi^4}{90}
$$