Question

$\underset{n\to \infty }{\lim }\left[\frac{n}{n^2+1}+\frac{n}{2^2+n^2}+\dots +\frac{1}{2n}\right]=$

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

Answer 1

Answer: (A)

Let
$S_n=\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\dots +\frac{1}{2n}$
$=\sum _{r=1}^n\frac{n}{n^2+r^2}=\sum _{r=1}^n\frac{1}{n\left(1+\frac{r^2}{n^2}\right)}$
Hence, $S=\underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n\frac{1}{1+\frac{r^2}{n^2}}$
$=\underset{0}{\overset{1}{\int }}\frac{dx}{1+x^2}={\left({\tan }^{-1}x\right)}_0^1=\frac{\pi }{4}$