Does the sum $\sum_{n=1, k=1} (n+k)^{-2}$ converge? If so how do I show it? I can show that $\sum_{n=1, k=1} (n+k)^{-3}$ converges by splitting up the sum into two sums, $\sum_{\ell}\sum_{n+k=\ell} (n+k)^{-3}$, which is bounded by $2\sum_\ell \ell^{-2}$.
Thanks.
The sum does not converge: $$\sum_{k,n}\frac{1}{(n+k)^2}>\sum_{n=1}^{\infty}\sum_{k=1}^n\frac{1}{(n+k)^2}>\sum_{n=1}^{\infty}\sum_{k=1}^n\frac{1}{(n+n)^2} = \sum_{n=1}^{\infty}\frac{1}{4n}.$$
You can compare it with the integral test:
$$\int_1^\infty \int_1^\infty\frac{1}{(x+y)^2}dxdy=\int_0^\infty \frac{1}{1+y}dy\longrightarrow\infty$$
Just a flavor.
Theorem:
$$\sum_{x\in\mathbb{N}^n}\frac{1}{x_1^{\alpha_1}+x_2^{\alpha_2}+\cdots+x_n^{\alpha_n}}<\infty\Longleftrightarrow\sum_{i}\frac{1}{\alpha_i}<1$$
Now for any natural $n$ and $k$, $(n+k)^2\leq2(n^2+k^2)$, so that $\sum_{n,k}(\frac{1}{n+k})^2\geq\frac{1}{2}\sum_{n,k}\frac {1}{n^2+k^2}=\infty$ since $\frac{1}{2}+\frac{1}{2}=1$
Edit： The theorem is a generalization of problem 18 from ItaMO (team match) 2019 final round:
Find the smallest natural $m$ such that $\sum_{\mathbb{N}^4}\frac{1}{a^2+b^3+c^8+d^m}$ converges. The answer was $25$.
To prove sufficiency, let $h=\sum\frac{1}{\alpha_i}<1$ and note $\alpha_i>1$ for each $i$, then use $AM-GM$ inequality and Tonelli's theorem: $$\sum_{\mathbb{N}^n}\frac{1}{x_1^{\alpha_1}+x_2^{\alpha_2}+\cdots+x_n^{\alpha_n}}\leq \frac{1}{h(\prod\alpha_i)^{1/h}}\sum_{\mathbb{N}^n}\frac{1}{x_1^{\alpha_1/h}x_2^{\alpha_2/h}\cdots x_n^{\alpha_n/h}}=$$ $$=\frac{1}{h(\prod\alpha_i)^{1/h}}\sum_{\mathbb{N}}\frac{1}{x_1^{\alpha_1/h}}\sum_{\mathbb{N}}\frac{1}{x_2^{\alpha_2/h}}\cdots\sum_{\mathbb{N}}\frac{1}{x_n^{\alpha_n/h}}<\infty$$
For necessity, note the following and change to spherical cooedinates $$\sum_{x\in\mathbb{N}^n}\frac{1}{x_1^{\alpha_1}+x_2^{\alpha_2}+\cdots+x_n^{\alpha_n}}>\int_{x_i\geq1}\frac{1}{x_1^{\alpha_1}+x_2^{\alpha_2}+\cdots+x_n^{\alpha_n}}d\lambda^n=$$ $$=\frac{2}{\prod\alpha_i}\int_{x_i\geq1}\frac{x_1^{2/\alpha_i-1}x_2^{2/\alpha_2-1}\cdots x_n^{2/\alpha_n-1}}{x_1^{2}+x_2^{2}+\cdots+x_n^{2}}d\lambda^n\geq$$ $$\geq\frac{2}{\prod\alpha_i}\int_S\int_{r\geq\sqrt{n}}\frac{r^{2h-n}\cdot r^{n-1}f(S)}{r^2}drdS=\frac{2}{\prod\alpha_i}\int_Sf(S)dS\int_{r\geq\sqrt{n}}r^{2h-3}dr$$ where $f(S)$ is a product of trigonometric functions from the Jacobian, so its integral converges, and the integral in $r$ converges only if $2h-3<-1\Longleftrightarrow h<1$.
Actually, the last convergence is an iff, so we may prove sufficiency also with spherical coordinates.
