Question

$\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2n}} + ..... + \frac{1}{\sqrt{n^{2} + (n - 1)n}} \right\rbrack$ is equal to

• A. $2 + 2\sqrt{2}$
• B. $2\sqrt{2} - 2$
• C. $2\sqrt{2}$
• D. 2

Answer 1

Answer: (B)

$2\sqrt{2} - 2$

Answer 2

$y = \lim_{n \rightarrow \infty}\left\lbrack \frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + .... + \frac{1}{\sqrt{n^{2} + (n - 1)n}} \right\rbrack$

⇒ $y = \lim_{n \rightarrow \infty}\left\lbrack \frac{1}{n} + \frac{1}{n\sqrt{1 + \frac{1}{n}}} + .... + \frac{1}{n\sqrt{1 + \frac{(n - 1)}{n}}} \right\rbrack$

⇒ $y = \frac{1}{n}\lim_{n \rightarrow \infty}\left\lbrack 1 + \frac{1}{\sqrt{1 + \frac{1}{n}}} + .... + \frac{1}{\sqrt{1 + \frac{(n - 1)}{n}}} \right\rbrack$

$y = \lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k = 1}^{n}\frac{1}{\sqrt{1 + \frac{(k - 1)}{n}}}$, Put $\frac{k - 1}{n} = x$ and $\frac{1}{n} = dx$

⇒ $y = \lim_{n \rightarrow \infty}\int_{0}^{\frac{n - 1}{n}}\frac{dx}{\sqrt{1 + x}}$ $= \lim_{n \rightarrow \infty}2\left\lbrack \sqrt{1 + x} \right\rbrack_{0}^{\left( \frac{n - 1}{n} \right)}$

⇒ $y = 2\lim_{n \rightarrow \infty}\left\lbrack \sqrt{\frac{2n - 1}{n}} - 1 \right\rbrack$ $= 2\lim_{n \rightarrow \infty}\sqrt{\frac{2n - 1}{n}} - 2$

⇒ $y = 2\lim_{n \rightarrow \infty}\sqrt{2 - \frac{1}{n}} - 2 = 2\sqrt{2} - 2$.