# 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 $H^2(\mathbb{T}_\pi^2)$.

• Wouldn't it work to bound the sum by $\Sigma \frac{1}{j^4}+\Sigma \frac{1}{k^4}$? – DBFdalwayse Sep 6 '13 at 7:05
• @DBF I think that would indeed be an upper bound, but how do you estimate $\sum \frac{1}{j^4+k^4} \leq \sum \frac{1}{j^4} + \sum \frac{1}{k^4} = 2\sum \frac{1}{k^4}$? Sorry if this is obvious! – mjb Sep 6 '13 at 7:14
• @kahen Yes, you are right about $(j,k) \in \mathbb{N}^2$, but octave actually suggests, that it still is an upper bound. – mjb Sep 6 '13 at 7:15
• Do you want to know the limit, or just to know if it converges? – DBFdalwayse Sep 6 '13 at 7:16
• @DBF Only the convergence. – mjb Sep 6 '13 at 8:00

You can use the inequality $2xy\le x^2+y^2$ (seen by expanding $(x-y)^2\ge0$) to get $2j^2k^2\le j^4+k^4$: $$\sum_{j,k=1}^\infty\frac1{j^4+k^4}\le\frac12\sum_{j=1}^\infty\sum_{k=1}^\infty\frac1{j^2}\frac1{k^2}=\frac12\Bigl(\sum_{j=1}^\infty\frac1{j^2}\Bigr)^2<\infty.$$

• Thank you! This is by far the easiest solution! – mjb Sep 6 '13 at 8:04

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).

• Thanks a lot! The last part is of independent interest to me! Thanks! – mjb Sep 6 '13 at 8:05

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.

• Thank you! This is another elegant way! – mjb Sep 6 '13 at 8:07

Straightforward application of comparison test: $$\sum_{j,k=1}^\infty \frac{1}{j^4+k^4} < \sum_{j=1}^\infty \frac{1}{j^4} = \frac{\pi^4}{90}$$

• I don't see how one would easily show that inequality. Note that this is not the sum of $\frac{1}{j^4 + k^4}$ for some constant $k$, we're summing over $k$ as well. – Arthur Sep 6 '13 at 7:19
• This isn't quite correct, since you're really comparing with $$\sum_{j, k} \frac{1}{j^4 + k^4} = \sum_j \sum_k \frac{1}{j^4 + k^4} < \sum_j \frac{\pi^4}{90} = \infty$$ – user61527 Sep 6 '13 at 7:21
• It should be clear that $\frac{1}{j^4+k^4} < \frac{1}{j^4}$ for any nonzero values of $j$ or of $k$ – Professor Mant Sep 6 '13 at 7:29
• Yes, but the first sum is summed over $\mathbb{N}\times\mathbb{N}$ and the second one over $\mathbb{N}$, which makes quite a difference. – Daniel R Sep 6 '13 at 7:34
• @ProfessorMant What we're saying is that we want the sum \begin{align} &\frac{1}{1^4 + 1^4} + \frac{1}{2^4 + 1^4} + \frac{1}{3^4 + 1^4} + \cdots \\\\ + &\frac{1}{1^4 + 2^4} + \frac{1}{2^4 + 2^4} + \frac{1}{3^4 + 2^4} + \cdots \\\\ + &\frac{1}{1^4 + 3^4} + \frac{1}{2^4 + 3^4} + \frac{1}{3^4 + 3^4} + \cdots \\\\ &\vdots \end{align} And so on. And while you have proven that any row or column of this sums to less than $\frac{\pi^4}{90}$, that isn't quite enough for the whole grid. – Arthur Sep 6 '13 at 8:26