Question

The value of ${lim}_{n\to \infty }[\frac{n}{1+n^2}+\frac{n}{4+n^2}+\frac{n}{9+n^2}+\cdots +\frac{1}{2n}]$ is equal to [Bihar CEE 1994]

• A.
  
• B.
  
• C. 1
• D. None of these

Answer 1

The correct option is B

We have, ${lim}_{n\to \infty }[\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots +\frac{1}{2n}]$
$={lim}_{n\to \infty }{\sum }_{r=1}^n\frac{n}{r^2+n^2}={lim}_{n\to \infty }{\sum }_{r=1}^n\frac{n}{n^2(1+\frac{r^2}{n^2})}$
$={lim}_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n\left(\frac{1}{1+\frac{r^2}{n^2}}\right)}={\int }_0^1\frac{dx}{1+x^2},$
$\{Applyingformula,{lim}_{n\to \infty }{\sum }_{r=1}^{n-1}\left\{f\left(\frac{r}{n}\right)\right\}\frac{1}{n}={\int }_0^{1}f\left(x\right)dx\}$
$=[ta{n}^{-1}x{]}_0^1=ta{n}^{-1}1-ta{n}^{-1}0=\frac{\pi }{4}$