$\lim\limits_{n \rightarrow \infty}\left\{\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{n}{n^2+n^2}\right\}$, is equal to
Answer & explanation
Correct answer: option 3
Let $S=\lim\limits_{n \rightarrow \infty}\left\{\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{n}{n^2+n^2}\right\}$
$\Rightarrow S=\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{n}{n^2+r^2}=\frac{1}{n} \lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^n \frac{1}{1+(r / n)^2}$
$\Rightarrow S=\int\limits_0^1 \frac{1}{1+x^2} d x=\left[\tan ^{-1} x\right]_0^1=\tan ^{-1} 1-\tan ^{-1} 0=\frac{\pi}{4}$