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$\lim _{n \rightarrow \infty}\left[\frac{n}{n^2+1}+\frac{n}{2^2+n^2}+\ldots+\frac{1}{2 n}\right]=$
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The correct answer is:
$\frac{\pi}{4}$
Let
$S_n=\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{1}{2 n}$
$=\sum_{r=1}^n \frac{n}{n^2+r^2}=\sum_{r=1}^n \frac{1}{n\left(1+\frac{r^2}{n^2}\right)}$
Hence, $S=\lim _{n \rightarrow \infty} S_n=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1+\frac{r^2}{n^2}}$
$=\int_0^1 \frac{d x}{1+x^2}=\left(\tan ^{-1} x\right)_0^1=\frac{\pi}{4}$
$S_n=\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{1}{2 n}$
$=\sum_{r=1}^n \frac{n}{n^2+r^2}=\sum_{r=1}^n \frac{1}{n\left(1+\frac{r^2}{n^2}\right)}$
Hence, $S=\lim _{n \rightarrow \infty} S_n=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1+\frac{r^2}{n^2}}$
$=\int_0^1 \frac{d x}{1+x^2}=\left(\tan ^{-1} x\right)_0^1=\frac{\pi}{4}$
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