Question

$\underset{n\to \infty }{\lim }\left(\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+\dots +\frac{1}{\sqrt{3n^2}}\right)$ equal to

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

Answer 1

Answer: (D)

$\underset{n\to \infty }{\lim }\left(\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+\dots +\frac{1}{\sqrt{3n^2}}\right)$
$=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\frac{1}{\sqrt{4-{\left(\frac{1}{n}\right)}^2}}+\frac{1}{\sqrt{4-{\left(\frac{2}{n}\right)}^2}}+\dots +\frac{1}{\sqrt{4-{\left(\frac{n}{n}\right)}^2}}\right]$
$=\underset{h\to 0}{\lim }h\left[\frac{1}{\sqrt{4-h^2}}+\frac{1}{\sqrt{4-(2h{)}^2}}+\dots +\frac{1}{\sqrt{4-(nh{)}^2}}\right]$
$=\underset{0}{\overset{1}{\int }}dx\sqrt{4-x^2}={\left[{\sin }^{-1}\frac{x}{2}\right]}_0^1$
$={\sin }^{-1}\left(\frac{1}{2}\right)-{\sin }^{-1}(0)=\frac{\pi }{6}$