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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.

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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 $L=2\ln(1+\sqrt 2)$.

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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 $2\operatorname{arsinh}(1)$.

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  • $\begingroup$ In English the notation for function inverse to $\sinh$ is ${\rm arsinh}$. $\endgroup$ – Adam Latosiński Jun 15 at 15:49
  • $\begingroup$ Thank you! Could you also please show me how you obtained the relation between $u_n$ and $v_n$? $\endgroup$ – John Bon Jun 15 at 15:49
  • $\begingroup$ John: just separate the cases $i=n$ and the cases $j=n$, which overlap exactly for the term $i=j=n$. $\endgroup$ – Mindlack Jun 15 at 15:54
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Hint:

$$\lim_{n \to \infty} \sum_{i,j=1}^{n} \frac{1}{n}\frac{1}{\sqrt{i^2+j^2}}$$

$$=\lim_{n \to \infty}\sum_{i,j=1}^{n} \frac{1}{n}\times \frac{1}{n} \frac{1}{\sqrt{\frac{i^2}{n^2}+\frac{j^2}{n^2}}}$$

$$= \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{x^2+y^2}} dxdy$$

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  • $\begingroup$ Thank you, but I said that I want a solution without double integrals. $\endgroup$ – John Bon Jun 15 at 15:07
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Hint: Turning into double integral gives $$ \int_{[0,1]^2}\frac{1}{r}\,\mathrm{d}\mathcal{L}^2(x,y)=\int_0^{\pi/2}\int_0^{\min(\sec\theta,\csc\theta)}\,\mathrm{d}r\,\mathrm{d}\theta $$

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  • $\begingroup$ Thank you, but I said that I want a solution without double integrals. $\endgroup$ – John Bon Jun 15 at 15:07
  • $\begingroup$ Why would you want to avoid double integrals? You have a double sum to start with... $\endgroup$ – user10354138 Jun 15 at 15:12
  • $\begingroup$ Because I am a high school student and I only learn single variable real analysis. $\endgroup$ – John Bon Jun 15 at 15:15
  • $\begingroup$ They can definitely be avoided, otherwise the problem wouldn't be in a book for high school students. $\endgroup$ – John Bon Jun 15 at 15:27

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