Question

Let $a_n={\int }_0^{\frac{\pi }{2}}(1-\sin t{)}^n\sin 2tdt$ and $\underset{n\to \infty }{lim}\sum _{n=0}^n\frac{a_n}{n}=\frac{1}{k}$. Find the value of $k$.

Answer 1

$a_n={\int }_0^{\frac{\pi }{2}}(1-\sin t{)}^n\sin 2tdt$
Let $1-\sin t=u\Rightarrow -\cos tdt=du$

$a_n=2{\int }_0^1{u}^n(1-u)du=2({\int }_0^1{u}^{n}du-{\int }_0^1{u}^{n+1}du)=2(\frac{1}{n+1}-\frac{1}{n+2})$

Now, $\frac{a_n}{n}=2(\frac{1}{n(n+1)}+\frac{1}{n(n+2)})$

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

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

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

$=\underset{n\to \infty }{lim}[\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}]$

which reduces to $\frac{1}{2}=\frac{1}{k}$

$\therefore k=2$