<s>[INST] 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 $<1$.

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} &= \sum_{k=1}^\infty \frac{1}{2k^4} + \sum_{j,k=1; j\neq k} \frac{1}{j^4+k^4} \\ &= \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}\\ &\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 $H^2(\mathbb{T}_\pi^2)$.
 [/INST] Grouping the terms by $i = (j + k)$, we get
$$
\sum_{i = 2}^\infty\:\: \sum_{j, k \gt 0; j + k = i} \frac{1}{j^4 + k^4}
$$
we can estimate that the inner sum, $\sum_{j, k \gt 0; j + k = i} \frac{1}{j^4 + k^4} \leq\frac{i - 1}{(i/2)^4}$, since we have $j^4 + k^4 \geq \left(\frac{j + k}{2}\right)^4 = (i/2)^4$. So we need to check whether the sum $$\sum_{i = 2}^\infty \frac{i-1}{(i/2)^4}$$
converges. But we see that 
$$
\frac{i-1}{(i/2)^4} = \frac{2(i - 1)}{i(i/2)^3} = \frac{2(1 - 1/i)}{(i/2)^3} = \frac{2 - 1/i}{i^3/8} \leq \frac{16}{i^3}
$$
for $i \geq 2$. This is easily seen to converge, and so the original series must converge.
</s>