Question

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

• A. $1/2$
• B. $1$
• C. $4/3$
• D. $3/2$

Answer 1

The correct option is D $3/2$

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

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

$-\cos tdt=du$

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

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

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

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

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

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

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

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

Calculating both the therms separately

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

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

$=(1-0)=1$

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

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

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

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