Question

The value of
$\underset{n\to \infty }{lim}[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{1}{2n}]$
is

• A. $\frac{n\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{4n}$
• D. $\frac{\pi }{2n}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$\underset{n\to \infty }{lim}[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{1}{2n}]\underset{n\to \infty }{lim}[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{n}{n^2+n^2}]$
$k^{th}$ term of the above series is
$T_k=\frac{n}{n^2+k^2}$

So, $[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{1}{2n}]=\sum _{k=1}^n\frac{n}{n^2+k^2}\therefore \underset{n\to \infty }{lim}[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{1}{2n}]=\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{n^2+k^2}\underset{n\to \infty }{lim}\sum _{k=1}^n\frac{n}{n^2+k^2}=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k^2}{n^2}}={\int }_0^1\frac{1}{x^2+1}dx=[{\tan }^{-1}x{]}_0^1=\frac{\pi }{4}$