Question

$\displaystyle\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{4 n^{2}-1}}+\frac{1}{\sqrt{4 n^{2}-2^{2}}}+\ldots+\frac{1}{\sqrt{3 n^{2}}}\right)$ equal to

• A. $0$
• B. $1$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (D)

$\displaystyle\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{4 n^{2}-1}}+\frac{1}{\sqrt{4 n^{2}-2^{2}}}+\ldots+\frac{1}{\sqrt{3 n^{2}}}\right)$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{\sqrt{4-\left(\frac{1}{n}\right)^{2}}}+\frac{1}{\sqrt{4-\left(\frac{2}{n}\right)^{2}}}+\ldots+\frac{1}{\sqrt{4-\left(\frac{n}{n}\right)^{2}}}\right]$
$=\displaystyle\lim _{h \rightarrow 0} h\left[\frac{1}{\sqrt{4-h^{2}}}+\frac{1}{\sqrt{4-(2 h)^{2}}}+\ldots+\frac{1}{\sqrt{4-(n h)^{2}}}\right]$
$=\int\limits_{0}^{1} d x \sqrt{4-x^{2}}=\left[\sin ^{-1} \frac{x}{2}\right]_{0}^{1}$
$=\sin ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}(0)=\frac{\pi}{6}$