Question

Let $a_n=\underset{0}{\overset{\pi /2}{\int }}(1-\sin t{)}^n\sin 2tdt$ then $\underset{n\to \infty }{\mathrm{Lim}}\sum _{n=1}^n\frac{a_n}{n}$ is equal to

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

Answer 1

Answer: (A)

$a_n=\underset{0}{\overset{\pi /2}{\int }}(1-\sin t{)}^n\sin 2tdt$
Let $1-\sin t=u\Rightarrow -\cos tdt=du$
$a_n=2\underset{0}{\overset{1}{\int }}{u}^n(1-u)du=2\left(\underset{0}{\overset{1}{\int }}{u}^{n}du-\underset{0}{\overset{1}{\int }}{u}^{n+1}du\right)=2\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
hence $\frac{a_n}{n}=2\left(\frac{1}{n(n+1)}-\frac{1}{n(n+2)}\right)$
$\underset{n\to \infty }{\mathrm{Lim}}\sum _{n=1}^n\frac{a_n}{n}=2\left(\sum \left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2}\sum \left(\frac{1}{n}-\frac{1}{n+2}\right)\right)=2\sum _{n=1}^n\left(\frac{1}{n}-\frac{1}{n+1}\right)-\sum _{n=1}^n\left(\frac{1}{n}-\frac{1}{n+2}\right)$
$=2(1)-\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\dots ..\right]=2-\frac{3}{2}=\frac{1}{2}$