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 $H^2(\mathbb{T}_\pi^2)$.
 A: You can use the inequality $2xy\le x^2+y^2$ (seen by expanding $(x-y)^2\ge0$) to get $2j^2k^2\le j^4+k^4$:
$$\sum_{j,k=1}^\infty\frac1{j^4+k^4}\le\frac12\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{j^2}\frac1{k^2}=\frac12\Bigl(\sum_{j=1}^\infty\frac1{j^2}\Bigr)^2<\infty.$$
A: 
Why does the series converge?

Because  the number of terms $N_n$ in the double sum such that $j^4+k^4\leqslant n$ is such that $N_n\leqslant c\sqrt{n}$ and because the double series is exactly
$$
\sum_{n\geqslant1}\frac1n(N_n-N_{n-1})=\sum_{n\geqslant1}\frac{N_n}{n(n+1)}\leqslant\sum_{n\geqslant1}\frac{c\sqrt{n}}{n^2},
$$
which converges.
To show the upper bound on $N_n$ note that the points $(j,k)$ such that $j^4+k^4\leqslant n$ are all in the (Euclidean) ball centered at $(0,0)$ with radius $R_n=\sqrt[4]{2n}$ since $(j^2+k^2)^2\leqslant2(j^4+k^4)$. Now, $N_n$ enumerates points in the first quadrant only, hence $N_n\leqslant\frac14\pi R_n^2\leqslant 1.12\cdot\sqrt{n}$.
The power of this approach is to reduce the determination of the (absolute) convergence of any positive double series 
$$
\sum\limits_{(j,k)\in\mathbb Z^2}\frac1{\varphi(j,k)}
$$ 
to an estimate of the number $N_n$ of indices $(j,k)$ such that $\varphi(j,k)\leqslant n$. As soon as the series $\sum\limits_{n\geqslant1}\frac{N_n}{n^2}$ converges, for example because $N_n=O(n^\alpha)$ with $\alpha\lt1$, one knows that the double series converges (and in fact this condition is necessary as well).
A: 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.
A: Straightforward application of comparison test: 
$$
\sum_{j,k=1}^\infty \frac{1}{j^4+k^4} < \sum_{j=1}^\infty \frac{1}{j^4} = \frac{\pi^4}{90}
$$
