Ltn→∞ 12n−12+14n−22+16n−32.....+1n=

ltn→∞12n−12+14n−22+......+1n=ltn→∞12n−12+14n−22+......+1n(2n)−n2=ltn→∞∑r=1n1(2n)r−r2=ltn→∞∑r=1n1n22rn−rn2=ltn→∞1n∑r=1n12rn−rn2=∫0112x−x2=∫0111−(x−1)2=(sin−1(x−1))01=sin−1(0)−sin−1(−1)=0+sin−11=π2

Ltn→∞ 12n−12+14n−22+16n−32.....+1n=

ltn→∞12n−12+14n−22+......+1n=ltn→∞12n−12+14n−22+......+1n(2n)−n2=ltn→∞∑r=1n1(2n)r−r2=ltn→∞∑r=1n1n22rn−rn2=ltn→∞1n∑r=1n12rn−rn2=∫0112x−x2=∫0111−(x−1)2=(sin−1(x−1))01=sin−1(0)−sin−1(−1)=0+sin−11=π2

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$\underset{n \to \infty}{L t} (\frac{1}{\sqrt{2 n - 1^{2}}} + \frac{1}{\sqrt{4 n - 2^{2}}} + \frac{1}{\sqrt{6 n - 3^{2}}} ..... + \frac{1}{n}) =$

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$\frac{π}{4}$

$\frac{π}{2}$

$\frac{π}{6}$

$\frac{π}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} \underset{n \to \infty}{l t} (\frac{1}{\sqrt{2 n - 1^{2}}} + \frac{1}{\sqrt{4 n - 2^{2}}} + ...... + \frac{1}{n}) \\ = \underset{n \to \infty}{l t} (\frac{1}{\sqrt{2 n - 1^{2}}} + \frac{1}{\sqrt{4 n - 2^{2}}} + ...... + \frac{1}{\sqrt{n (2 n) - n^{2}}}) \\ = \underset{n \to \infty}{l t} \sum_{r = 1}^{n} \frac{1}{\sqrt{(2 n) r - r^{2}}} \\ = \underset{n \to \infty}{l t} \sum_{r = 1}^{n} \frac{1}{\sqrt{n^{2} [2 (\frac{r}{n}) - {(\frac{r}{n})}^{2}]}} \\ = \underset{n \to \infty}{l t} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{\sqrt{2 (\frac{r}{n}) - {(\frac{r}{n})}^{2}}} \\ = \int_{0}^{1} \frac{1}{\sqrt{2 x - x^{2}}} \\ = \int_{0}^{1} \frac{1}{\sqrt{1 - {(x - 1)}^{2}}} \\ = {(\sin^{- 1} (x - 1))}_{0}^{1} = \sin^{- 1} (0) - \sin^{- 1} (- 1) \\ = 0 + \sin^{- 1} 1 = \frac{π}{2} \end{array}$