Compute $\lim \limits_{n\to \infty} \frac{1}{n}\sum_{i,j=1}^n\frac{1}{\sqrt{i^2+j^2}}$ Compute $\lim \limits_{n\to \infty} \frac{1}{n}\sum_{i,j=1}^n\frac{1}{\sqrt{i^2+j^2}}$.
I am not looking for a solution using double integrals. I tried to turn this into a Riemann sum, but I couldn't make any progress.
I thought about using the squeeze theorem, but I can't find any useful inequalities, I just tried to use AM-GM on the denominator, but it didn't help.
Edit: This is definitely solveable without double integrals, it comes from a high school book and here multivariate calculus isn't covered. 
 A: Denote the limit you have to compute by $L$.
We may apply the Stolz-Cesaro theorem to get that $L=\lim\limits_{n\to \infty}\left(2\sum_{i=1}^{n-1}\frac{1}{\sqrt{n^2+i^2}}+\frac{1}{\sqrt{2n^2}}\right)$.
Since $\lim\limits_{n\to \infty}\frac{1}{\sqrt{2n^2}}=0$ and $\lim\limits_{n\to \infty}\sum_{i=1}^{n-1}\frac{1}{\sqrt{n^2+i^2}}=\int\limits_0^1\frac{dx}{\sqrt{x^2+1}}=\ln(1+\sqrt 2)$, we get that $L=2\ln(1+\sqrt 2)$.
A: Define $$u_n=\sum_{\substack{1 \leq i,j \leq n \\ i=n \text { or } j=n}}{(i^2+j^2)^{-1/2}}.$$
You want to find $\lim \, \frac{1}{n}\sum_{k=1}^n{u_k}$. 
It is, provided that $\{u_n\}$ converges, the limit of $\{u_n\}$. 
Now, $u_n=2v_n-(\sqrt{2}n)^{-1}$ where $v_n=\sum_{k=1}^n{\frac{1}{\sqrt{n^2+k^2}}}$. 
Now, $v_n=\frac{1}{n} \sum_{k=1}^n{(1+(k/n)^2)^{-1/2}} \rightarrow \int_0^1{\frac{dx}{\sqrt{x^2+1}}}=\operatorname{arsinh}(1)$. 
Thus your limit is $2\operatorname{arsinh}(1)$. 
A: Hint: 
$$\lim_{n \to \infty} \sum_{i,j=1}^{n} \frac{1}{n}\frac{1}{\sqrt{i^2+j^2}}$$
$$=\lim_{n \to \infty}\sum_{i,j=1}^{n} \frac{1}{n}\times \frac{1}{n} \frac{1}{\sqrt{\frac{i^2}{n^2}+\frac{j^2}{n^2}}}$$
$$= \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{x^2+y^2}} dxdy$$
A: Hint: Turning into double integral gives
$$
\int_{[0,1]^2}\frac{1}{r}\,\mathrm{d}\mathcal{L}^2(x,y)=\int_0^{\pi/2}\int_0^{\min(\sec\theta,\csc\theta)}\,\mathrm{d}r\,\mathrm{d}\theta
$$
