About the convergence of $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k^{2}+n^{1/\gamma}}$ Does the series converge?
$$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k^{2}+n^{1/\gamma}}$$
 A: Hint:
$$
\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^2 + n^{1/\gamma}} \geq \sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2 + n^2}.
$$
A: The summand in the double sum are all positive numbers. If the double sum converges, it converges absolutely and we can evaluate the result in any order we want.
Notice for any $x\in \mathbb{R}_{+}$,
$$\sum_{k=1}^{\infty}\frac{1}{k^2+x^2} = \frac{\pi}{2x}\left[\frac{\cosh(\pi x)}{\sinh(\pi x)}-\frac{1}{\pi x}\right]$$
If the double sum converges, we have:
$$\begin{align}
\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k^2+n^{1/\gamma}}
\stackrel{?}{=} & \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k^2+n^{1/\gamma}}\\
= & \sum_{n=1}^{\infty} \frac{\pi}{2 n^{1/(2\gamma)}} \left[
\frac{\cosh(\pi n^{1/(2\gamma)})}{\sinh(\pi n^{1/(2\gamma)})}-\frac{1}{\pi n^{1/(2\gamma)}}\right]\\
= & \sum_{n=1}^{\infty} \frac{\pi}{2 n^{1/(2\gamma)}} \left[
\frac{\cosh(\pi n^{1/(2\gamma)})}{\sinh(\pi n^{1/(2\gamma)})}\right] - \frac12\zeta(\frac{1}{\gamma})\tag{*}
\end{align}$$
As $n \to \infty$, $\coth(\pi n^{1/(2\gamma)}) \sim 1 + O(e^{-2\pi n^{1/(2\gamma)}})$. The partial sums in the of R.H.S of $(*)$ behaves like:
$$\sum_{n=1}^N \frac{\pi}{2 n^{1/(2\gamma)}} + O(1) \sim \frac{\pi}{2}\int_{1}^N x^{-1/(2\gamma)} dx + O(1) \sim \frac{\pi\gamma}{2\gamma-1} N^{1-1/(2\gamma)} + O(1)$$
Since $1 - 1/(2\gamma) > 0$, $(*)$ diverges. This contradict with our original assumption that the double sum converges. So the original double sum diverges.
