Question

Let $a_n={\int }_0^{\frac{\pi }{2}}{(1-\sin t)}^n\sin 2tdt$ then ${lim}_{n\to \infty }{\sum }_{n=1}^n\frac{a_n}{n}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{1}{3}$
• D. $\frac{4}{2}$

Answer 1

Answer: (A), (B)

$\frac{1}{2}$
$\frac{3}{2}$

$a_n={\int }_0^{\pi /2}{(1-\sin t)}^n\sin 2tdt$

Let $1-\sin t=u\Rightarrow \sin t$

$-\cos tdt=du$

$={\int }_1^0{u}^n\left(2(u-1)\right)du$

$=2{[\frac{u^{n+2}}{n+2}-\frac{u^{n+1}}{n+1}]}_1^0$

$=2(\frac{1}{n+1}-\frac{1}{n+2})=2\left(\frac{1}{(n+1)(n+2)}\right)$

$L={lim}_{n\to \infty }{\sum }_{n=1}^n\frac{a_n}{n}={lim}_{n\to \infty }{\sum }_{n=1}^n\frac{2}{n(n+1)(n+2)}$

$={lim}_{n\to \infty }{\sum }_{n=1}^n\frac{(n+2)-n}{n(n+1)(n+2)}$

$={lim}_{n\to \infty }{\sum }_{n=1}^n(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})$

$={lim}_{n\to \infty }{\sum }_{n=1}^n\frac{(n+1)-n}{n(n+1)}-{lim}_{n\to \infty }{\sum }_{n=1}^n\frac{(n+2)-(n+1)}{(n+1)(n+2)}$

$={lim}_{n\to \infty }[\left({\sum }_{n=1}^n(\frac{1}{n}-\frac{1}{n+1})\right)-\left({\sum }_{n=1}^n(\frac{1}{n+1}-\frac{1}{n+2})\right)]$

Calculating both the therms separately

$1.{lim}_{n\to \infty }(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n}-\frac{1}{n+1})$

$={lim}_{n\to \infty }(1-\frac{1}{n+1})$ as $n\to \infty ,\frac{1}{n}\to 0$

$=(1-0)=1$

$2.{lim}_{n\to \infty }(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n+1}-\frac{1}{n+2})$

$={lim}_{n\to \infty }(\frac{1}{2}-\frac{1}{n+2})$

$=\frac{1}{2}-0=\frac{1}{2}$

$L=1+\frac{1}{2}=\frac{3}{2}$