Question

The value of $\lim_{n \rightarrow \infty}\left\lbrack \frac{n}{1 + n^{2}} + \frac{n}{4 + n^{2}} + \frac{n}{9 + n^{2}} + .... + \frac{1}{2n} \right\rbrack$is equal to

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. 1
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi}{4}$

Answer 2

We have, $\lim_{n \rightarrow \infty}\left\lbrack \frac{n}{1 + n^{2}} + \frac{n}{4 + n^{2}} + ...... + \frac{1}{2n} \right\rbrack$

$= \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}\frac{n}{r^{2} + n^{2}} = \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}\frac{n}{n^{2}\left( 1 + \frac{r^{2}}{n^{2}} \right)}$

$= \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}{\frac{1}{n\left( 1 + \frac{r^{2}}{n^{2}} \right)} = \int_{0}^{1}\frac{dx}{1 + x^{2}}}$,

$\left\{ \text{Applying formula, }\lim_{n \rightarrow \infty}\sum_{r = 0}^{n - 1}{\left\{ f\left( \frac{r}{n} \right) \right\}.\frac{1}{n} = \int_{0}^{1}{f(x)dx}} \right\}$

$= \lbrack\tan^{- 1}x\rbrack_{0}^{1} = \tan^{- 1}1 - \tan^{- 1}0 = \frac{\pi}{4}.$