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Is the series $$ \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty} {1 \over \left(k^{2} + n^{2}\right)^{\left(1+\epsilon\right)/2}} $$ convergent for $\epsilon > 0$ ? I don't know how to manage the double sum.

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The series converges if $\epsilon>1$. Since $2kn\leq k^2+n^2$, \begin{align*} \sum_{n=1}^\infty\sum_{k = 1}^\infty{1\over(\sqrt{n^2+k^2})^{1+\epsilon}}&\leq {1\over 2^{(1+\epsilon)/2}}\sum_{n=1}^\infty{1\over n^{(1+\epsilon)/2}}\sum_{k = 1}^\infty{1\over k^{(1+\epsilon)/2}} \\ & = {1\over 2^{(1+\epsilon)/2}}\left(\sum_{n=1}^\infty{1\over n^{(1+\epsilon)/2}}\right)^2, \end{align*} and that last series converges for $\epsilon>1$. On the other hand, $k^2+n^2\leq (k+n)^2$, and so \begin{align*} \sum_{n=1}^\infty\sum_{k = 1}^\infty{1\over(\sqrt{n^2+k^2})^{1+\epsilon}} &\geq \sum_{n=1}^\infty\sum_{k = 1}^\infty{1\over(n+k)^{1+\epsilon}} \\ &= \sum_{r=1}^\infty {|\{(n,k):n+k=r\}|\over r^{1+\epsilon}} \\ & = \sum_{r=1}^\infty {r-1\over r^{1+\epsilon}} \end{align*} (The last equality follows from the fact that there are $r-1$ ordered pairs $(n,k)$ such that $n+k=r$, namely, $(n,k)=(j,r-j)$ for $j=1,\dots,r-1$.) The final sum diverges for $\epsilon\leq 1$, hence the initial one does as well.

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  • $\begingroup$ Ok all fixed.${}{}{}{}$ $\endgroup$ – Nick Strehlke Jan 7 '14 at 20:13
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The problem is equivalent to the convergence of $\sum_{k,l}(k+l)^{-(1+\varepsilon)}.$ Using an estimation of the remainder $\sum_{j\geqslant r}j^{-(1+\varepsilon)}$ by $r^{-\varepsilon}$, we can see that the initial double series is convergent if and only if $\varepsilon \gt 1$.

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