Question

The value of $\underset{n\to \infty }{lim}[\frac{1}{\sqrt{(2n-1^2)}}+\frac{1}{\sqrt{(4n-2^2)}}+\frac{1}{\sqrt{(6n-3^2)}}+...+\frac{1}{n}]$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{\pi }{2}$

Let $P=\underset{n\to \infty }{lim}[\frac{1}{\sqrt{(2n-1^2)}}+\frac{1}{\sqrt{(4n-2^2)}}+\frac{1}{\sqrt{(6n-3^2)}}+\cdots +\frac{1}{n}]$
$=\underset{n\to \infty }{lim}[\frac{1}{\sqrt{1\left(2n\right)-1^2}}+\frac{1}{\sqrt{2\left(2n\right)-2^2}}+\frac{1}{\sqrt{3\left(2n\right)-3^2}}+\cdots +\frac{1}{\sqrt{n\left(2n\right)-n^2}}]=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{\sqrt{r\left(2n\right)-r^2}}=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n\cdot \sqrt{2\frac{r}{n}-{\left(\frac{r}{n}\right)}^2}}=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{(2x-x^2)}}$
Put $x=t^2\Rightarrow dx=2tdt$
$=\underset{0}{\overset{1}{\int }}\frac{2tdt}{t\sqrt{2-t^2}}={\left[2{\sin }^{-1}\left(\frac{t}{\sqrt{2}}\right)\right]}_0^1=2{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)=2\left(\frac{\pi }{4}\right)=\frac{\pi }{2}$