Question

$\underset{n\to \infty }{lim}[\frac{1}{2n}+\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-4}}+....+\frac{1}{\sqrt{3n^2+2n-1}}]$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is D $\frac{\pi }{6}$

Sol:

$S={\sum }_{k=0}^{n-1}\frac{1}{\sqrt{{4n}^2-k^2}}={\sum }_{k=0}^{n-1}\frac{1}{\sqrt{4-{\left(\frac{k}{n}\right)}^2}}\frac{1}{n}$

We can write the above summation in the form of intergral as

$S={\sum }_{k=0}^{n-1}\frac{1}{\sqrt{{4n}^2-k^2}}={\sum }_{k=0}^{n-1}\frac{1}{\sqrt{4-{\left(\frac{k}{n}\right)}^2}}\frac{1}{n}$

Integrating the above, we get

$S={\sin }^{-1}{\left[\left(\frac{x}{2}\right)\right]}_0^1={\sin }^{-1}\left(\frac{1}{2}\right)-{\sin }^{-1}\left(0\right)=\frac{\pi }{6}$