Calculate $$\lim_{n\to\infty}\sum_{\ \ \ \ i,j\ge0 \\ i^2+j^2\le n^2} \frac{1}{n^2+i^2+j^2}$$
It smells like a Riemann sum for a double integral. $$\lim_{n\to\infty}\sum_{\ \ \ \ i,j\ge0 \\ i^2+j^2\le n^2} \frac{1}{n^2+i^2+j^2} = \lim_{n\to\infty}\frac{1}{n^2}\sum_{\ \ \ \ i,j\ge0 \\ i^2+j^2\le n^2} \frac{1}{1+(i/n)^2+(j/n)^2} \overset{?}= \iint\limits_{[0,1]^2} \frac{1}{1+x^2+y^2}\mathrm dx\mathrm dy$$ I know how to calculate the last integral, but I'm not sure if the (?) part is correct. Doesn't $i^2+j^2\le n$ have any special job to do here?
The limit is $$\iint\limits_{\substack{x,y\geqslant 0\\x^2+y^2\leqslant 1}}\frac{{\rm d}x\,{\rm d}y}{1+x^2+y^2}=\iint\limits_{\substack{0\leqslant r\leqslant 1\\0\leqslant\phi\leqslant\pi/2}}\frac{r\,{\rm d}r\,{\rm d}\phi}{1+r^2}=\frac{\pi\log 2}{4}.$$