Question

The value of $\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\sqrt{\frac{n+r}{n-r}}$ is

• A. $\frac{\pi }{2}$
• B. $2\pi$
• C. $\frac{\pi }{2}-1$
• D. $\frac{\pi }{2}+1$

Answer 1

The correct option is D $\frac{\pi }{2}+1$

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\sqrt{\frac{n+r}{n-r}}$

Converting limit into integral, we get

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\sqrt{\frac{1+\frac{r}{n}}{1-\frac{r}{n}}}={\int }_0^1\sqrt{\frac{1+x}{1-x}}dx$
$x=\cos \theta \to dx=-\sin \theta d\theta$
${\int }_{\frac{\pi }{2}}^0\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}(-\sin \theta )d\theta$
$={\int }_0^{\frac{\pi }{2}}(1-1+2{\cos }^2\frac{\theta }{2})d\theta$

$={\int }_0^{\frac{\pi }{2}}(2{\cos }^2\frac{\theta }{2}-1)d\theta +{\int }_0^{\frac{\pi }{2}}d\theta$

$={\int }_0^{\frac{\pi }{2}}\cos \theta d\theta +\frac{\pi }{2}$

$=1+\frac{\pi }{2}$

Hence, option D.