Question

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

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

Answer 1

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

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

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

Now let $x=\frac{r}{n}$, and hence when $n\to \infty$, $x=\frac{r}{n}\to 0$

Hence,

${\int }_0^1\sqrt{\frac{1+x}{1-x}}dx$

Now we have to integrate this-

Let $x=\sin \theta$

$⟹dx=\cos \theta d\theta$

And , at $x=1,\theta =\frac{\pi }{2}$ and for $x=0,\theta =0$

Hence, integration becomes,

${\int }_0^{\frac{\pi }{2}}\sqrt{\frac{1+sin\theta }{1-sin\theta }}cos\theta d\theta$

$={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{1+sin\theta }}{\sqrt{1-sin\theta }}\times \sqrt{\frac{1+sin\theta }{1+sin\theta }}cos\theta d\theta$

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

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

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

$={[\theta -cos\theta ]}_0^{\frac{\pi }{2}}$

$={\left[\theta \right]}_0^{\frac{\pi }{2}}-{\left[cos\theta \right]}_0^{\frac{\pi }{2}}$

$=(\frac{\pi }{2}-0)-(\cos \frac{\pi }{2}-\cos 0)$

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

Hence, correct answer is option-(D).