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I need to prove that

$$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2 + k^2} = \frac{\pi}{4}$$

using definite integrals. I would like to get a explanation about to do it , and whats the reasons. thanks!!

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    $\begingroup$ and what did you try? $\endgroup$ – alexjo Mar 31 '16 at 15:17
  • $\begingroup$ @alexjo Well i thought about trying to convert it to a function and try to calculate the integral. but just got the feeling I'm not in the right way. $\endgroup$ – Barak michaeli Mar 31 '16 at 15:19
  • $\begingroup$ You have to recognize this as a Riemann sum for an appropriate function. The limit is then the actual integral. $\endgroup$ – Paul Mar 31 '16 at 15:22
  • $\begingroup$ $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1+(k/n)^2} = \int_0^1 \frac{dx}{1+x^2}$$ $\endgroup$ – Crostul Mar 31 '16 at 15:22
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$$\lim _{n\rightarrow \infty }\sum_{k=1}^{n}\frac{n}{n^{2}+k^{2}}=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\left(\frac{k}{n}\right)^2}=\int_{0}^{1}\frac{dx}{1+x^2}=\arctan 1=\color{red}{\frac{\pi}{4}}.$$

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