The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$ [closed]

How to compute this limit:

$$\lim_{n\to\infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+n^2}\right)$$

closed as off-topic by choco_addicted, Chinnapparaj R, A. Pongrácz, RRL, José Carlos SantosDec 5 '18 at 14:35

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\begin{align}\lim_{n\to\infty}\sum_{1\le r\le n}\frac{n}{n^2+r^2} &=\lim_{n\to\infty}\frac1n\sum_{1\le r\le n}\frac1{1+\left(\frac rn\right)^2}\\ &=\int_0^1\frac{dx}{1+x^2}\end{align}
$$\text{as }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
• @lab bhattacharjee. You provided, for sure, the best and simplest solution to the problem. What amazed me was to have a look to the sum up to $n$. What I got (using a CAS) is $$\frac{-i n \left(H_{(1-i) n}-H_{(1+i) n}\right)+\pi n \coth (\pi n)-1}{2 n}$$ which goes to the limit you gave (fortunately for me !). Cheers. – Claude Leibovici Jul 19 '14 at 9:14