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

Answer: (B)

$\frac{\pi }{4}$

Let $L=\underset{n\to \infty }{lim}[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+.....+\frac{1}{2n}]$

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

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

$=\underset{n\to \infty }{lim}\frac{1}{n}[\frac{1}{1+(\frac{1}{n}{)}^2}+\frac{1}{1+(\frac{2}{n}{)}^2}+.....+\frac{1}{1+(\frac{n}{n}{)}^2}]$

$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\frac{1}{1+{\left(\frac{r}{n}\right)}^2}$

Take $x=\frac{r}{n}$, as $r=1⟹x=\frac{1}{n}\to 0\because n\to \infty$

Also, for $r=n,x=\frac{n}{n}=1$

So, we get the limit from $0$ to $1$

$\therefore L={\int }_0^1\frac{1}{1+x^2}dx$

$=|{\tan }^{-1}x{|}_0^1=\frac{\pi }{4}$

Hence, $\underset{n\to \infty }{lim}[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+.....+\frac{1}{2n}]=\frac{\pi }{4}$