TITLE: Proving that $\sum_{(m,n)\in \Bbb Z \times \Bbb Z}\frac{1}{m^2+n^2+1}$ diverges.
QUESTION [11 upvotes]: I must prove that $$\sum_{(m,n)\in \Bbb Z \times \Bbb Z}\frac{1}{m^2+n^2+1}$$ diverges. I understand that given $K > 0$, I must find a finite subset of $\Bbb Z \times \Bbb Z$, say, $F$, such that $\sum_{(m,n)\in F}\frac{1}{m^2+n^2+1} > K$.
I know that I can't be lazy and just walk in diagonals, because $\sum 1/(2n^2+1) < +\infty$.
Call $Q_n = ([-n,n]\times[-n,n])\cap (\Bbb Z \times \Bbb Z)$. Sure, $Q_n$ is finite for all $n$. Given $K > 0$, I want to find $n$ such that the sum over $Q_n$ goes over $K$. I can't think of a better choice of finite set to compute the sum over: I'm going in all directions, certainly it should blow up. For example, $Q_2$:

Using and abusing of symmetry, I got so far (hopefully correct): $$ \sum_{(a,b)\in Q_n}\frac{1}{a^2+b^2+1} = 1+4\sum_{i=1}^n\frac{1}{i^2+1} +4\sum_{i=1}^n\frac{1}{2i^2+1} +8\sum_{1\leq i<j\leq n}\frac{1}{i^2+j^2+1}.$$
Needless to say, this is too complicated of an expression for me to find $n = n(K)$ to do the job - I don't know even how to find closed expressions for each term above. Maybe using $Q_n$ is overkill, and we can find a smaller set to compute the sum over.
Hints, ideas, a magic solution? Thanks!

REPLY [2 votes]: Compute a lower bound for the summation in $j$ for each $i$ using
$$
\begin{align}
\sum_{j\in\mathbb{Z}}\frac1{i^2+j^2+1}
&\ge\overbrace{\vphantom{\int}\frac1{i^2+1}}^{j=0}+\overbrace{2\int_1^\infty\frac{\mathrm{d}x}{i^2+x^2+1}}^{j\ge1\text{ and }j\le-1}\\
&=\frac1{i^2+1}+\frac2{\sqrt{i^2+1}}\int_{\frac1{\sqrt{i^2+1}}}^\infty\frac{\mathrm{d}x}{x^2+1}\\
&\ge\frac1{i^2+1}+\frac2{\sqrt{i^2+1}}\int_1^\infty\frac{\mathrm{d}x}{x^2+1}\\
&\ge\frac1{i^2+1}+\frac{\pi/2}{\sqrt{i^2+1}}\\
\end{align}
$$
Now show that this series diverges when summed in $i$ by comparing with the harmonic series.