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The series $$\sum_{k=1}^\infty \zeta (2k)z^{2k}$$ (where $\zeta$ is the zeta function) has a non-zero radius of convergence (in $z$). This led me to the following question:

Let $$C_k=\sum_{m=1}^\infty\sum_{n=0}^\infty \frac{1}{(m^2+n^2)^{2k}}$$ where $k$ is an integer $\ge 1$. Does $$\sum_{k=1}^\infty C_k z^{4k}$$ have a non-zero radius of convergence as well?

In fact, $C_k$ is "almost" the Eisenstein series (https://en.wikipedia.org/wiki/Eisenstein_series) of weight $4k$ evaluated at $i$ (known to converge absolutely), with the summand replaced by absolute value: $$C_k=\sum_{m=1}^\infty\sum_{n=0}^\infty \frac{1}{|m+ni|^{4k}}.$$

I tried to build on the analogy with the zeta function. Using the fact that $m^2+n^2\ge n^2$, I obtained $$C_k\le \zeta (4k)+\sum_{m=1}^\infty \zeta (4k)$$ but, unfortunately, this doesn't lead anywhere.

Nomas2
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1 Answers1

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Perhaps the easiest approach is to note that $C_k$ is decreasing in $k$. As $C_1$ is finite, we have $$ \left \lvert \sum_{k \geq 1} C_k z^k \right \rvert \leq \sum_{k \geq 1} \lvert C_k \rvert \lvert z \rvert^k \leq C_1 \sum_{k \geq 1} \lvert z \rvert^k \leq C_1 \frac{\lvert z \rvert}{1 - \lvert z \rvert},$$ assuming that $\lvert z \rvert < 1$ so that the geometric series converges.

davidlowryduda
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