$lim_{n \to \infty} [\frac{1}{\sqrt{(n^{2} - 1)}} + \frac{1}{\sqrt{(n^{2} - 2^{2})}} + \dots \dots + \frac{1}{\sqrt{(n^{2} - {(n - 1)}^{2})}}]$

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$π$

$\frac{π}{2}$

none

answer is B.

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Detailed Solution

$lim_{n \to \infty} [\frac{1}{\sqrt{(n^{2} - 1)}} + \frac{1}{\sqrt{(n^{2} - 2^{2})}} + \dots \dots + \frac{1}{\sqrt{(n^{2} - {(n - 1)}^{2})}}]$

$= lim_{n \to \infty} \frac{1}{n} [\frac{1}{\sqrt{(1 - \frac{1}{n^{2}})}} + \frac{1}{\sqrt{[1 - {(\frac{2}{n})}^{2}]}} + \dots . .$ $+ \frac{1}{\sqrt{[1 - {(\frac{n - 1}{n})}^{2}]}}]$

$= lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n - 1} \frac{1}{\sqrt{[1 - {(\frac{r}{n})}^{2}]}}$

Replace $\frac{r}{n}$ by x and $\frac{1}{n}$ by dx.

Also when $r = 1, x = \frac{1}{n} = 0$

when $r = n - 1, x = \frac{n - 1}{n} = 1 - \frac{1}{n} = 1$

$\begin{array}{l} ∴ lim = \int_{0}^{1} \frac{1}{\sqrt{(1 - x^{2})}} d x = [\sin^{- 1} x]_{0}^{1} \\ = \sin^{- 1} 1 - \sin^{- 1} 0 = π / 2 \end{array}$