Question

$\displaystyle\lim _{n \rightarrow \infty}\left[\frac{n}{n^{2}+1}+\frac{n}{2^{2}+n^{2}}+\ldots+\frac{1}{2 n}\right]=$

• A. $\frac{\pi}{4}$
• B. log 2
• C. 0
• D. 1

Answer 1

Answer: (A)

Let
$S_{n}=\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\ldots+\frac{1}{2 n}$
$=\displaystyle\sum_{r=1}^{n} \frac{n}{n^{2}+r^{2}}=\displaystyle\sum_{r=1}^{n} \frac{1}{n\left(1+\frac{r^{2}}{n^{2}}\right)}$
Hence, $S=\displaystyle\lim _{n \rightarrow \infty} S_{n}=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \displaystyle\sum_{r=1}^{n} \frac{1}{1+\frac{r^{2}}{n^{2}}}$
$ =\int\limits_{0}^{1} \frac{d x}{1+x^{2}}=\left(\tan ^{-1} x\right)_{0}^{1}=\frac{\pi}{4}$