Intriguing Sum

$Consider:\quad I(n)=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2n }{ x } } dx\\ \\ \Rightarrow I(n)=\frac { 2n-1 }{ 2n } I(n-1)\quad ,\quad I(0)=\quad \frac { \pi }{ 2 } \\ \\ \Rightarrow I(n)\quad =\frac { (2n-1)(2n-3)(2n-5)....(1) }{ (2n)(2n-2)(2n-4)...(2) } \times \frac { \pi }{ 2 } \\ \\ \Rightarrow I(n)\quad =\left( \begin{matrix} 2n \\ n \end{matrix} \right) \times \frac { \pi }{ 2 } \times { 2 }^{ -2n }\quad \\ \Rightarrow \frac { \left( \begin{matrix} 2n \\ n \end{matrix} \right) }{ { n\times 4 }^{ n } } =\frac { 2 }{ \pi } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2n }{ x } }{ n } } dx\\ Consider:\quad -\log { (1-t) } =t+\frac { { t }^{ 2 } }{ 2 } +\frac { { t }^{ 3 } }{ 3 } +\frac { { t }^{ 4 } }{ 4 } +...\\ \Rightarrow -\log { (1-\sin ^{ 2 }{ t } ) } =\sin ^{ 2 }{ t } +\frac { { (\sin ^{ 2 }{ t } ) }^{ 2 } }{ 2 } +\frac { { (\sin ^{ 2 }{ t } ) }^{ 3 } }{ 3 } +\frac { { (\sin ^{ 2 }{ t } ) }^{ 4 } }{ 4 } +...\\ \Rightarrow \sum _{ n=1 }^{ \infty }{ \frac { \left( \begin{matrix} 2n \\ n \end{matrix} \right) }{ { n\times 4 }^{ n } } } =\frac { 2 }{ \pi } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\sin ^{ 2 }{ t } +\frac { { (\sin ^{ 2 }{ t } ) }^{ 2 } }{ 2 } +\frac { { (\sin ^{ 2 }{ t } ) }^{ 3 } }{ 3 } +\frac { { (\sin ^{ 2 }{ t } ) }^{ 4 } }{ 4 } +... } )dt\\ \Rightarrow \sum _{ n=1 }^{ \infty }{ \frac { \left( \begin{matrix} 2n \\ n \end{matrix} \right) }{ { n\times 4 }^{ n } } } =S=\frac { 2 }{ \pi } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ -\log { (1-\sin ^{ 2 }{ t } ) } } dt\\ \Rightarrow S=\frac { 2 }{ \pi } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ -2\log { (\cos { t } ) } } dt=\log { 4 } \\ \Rightarrow { e }^{ S }=\boxed { 4 }$

Here's another way Let ${ \left( 1+mx \right) }^{ n }$ = $1+\sum _{ n=1 }^{ \infty }{ \frac { \left( \begin{matrix} 2n \\ n \end{matrix} \right) }{ { 4 }^{ n } } }$ ${ x }^{ n }$

comparing coefficents of x and ${ x }^{ 2 }$

mn=1/2 and ${m}^{2}$ n(n-1)/2=1.3/2.4

solving we get m=-1,n=-1/2,we can easily check if other terms are also as in the first expression

So $\frac { 1 }{ { (1-x) }^{ 1/2 } } =1+\frac { \left( \begin{matrix} 2 \\ 1 \end{matrix} \right) x }{ { 4 }^{ 1 } } +\frac { \left( \begin{matrix} 4 \\ 2 \end{matrix} \right) { x }^{ 2 } }{ { 4 }^{ 2 } } +.....$

Taking 1 to the other side , and dividing by x and then integrating from x=0 to x=1 we get,

$\int _{ 0 }^{ 1 }{ \frac { 1 }{ { x(1-x) }^{ 1/2 } } - } \frac { 1 }{ x }dx =\int _{ 0 }^{ 1 }\frac { \left( \begin{matrix} 2 \\ 1 \end{matrix} \right) }{ { 4 }^{ 1 } } +\frac { \left( \begin{matrix} 4 \\ 2 \end{matrix} \right) { x }^{ 1 } }{ { 4 }^{ 2 } } +..... dx$

On performing this integration ,on we get,

$log(4)=\frac { \left( \begin{matrix} 2 \\ 1 \end{matrix} \right) }{ { 4 }^{ 1 } } .\frac { 1 }{ 1 } +\frac { \left( \begin{matrix} 4 \\ 2 \end{matrix} \right) }{ { 4 }^{ 2 } } .\frac { 1 }{ 2 } +.....$ which is equal to s

Hence e^s=4

In the series, $\left( \begin{matrix} 2n \\ n \end{matrix} \right)$ is the central binomial coefficient. However there are no exact formula of $\left( \begin{matrix} 2n \\ n \end{matrix} \right)$ , but we can use approximations. One good approximation is by using more terms in the Stirling series. We get

$\left( \begin{matrix} 2n \\ n \end{matrix} \right) =\frac { 4^{ n } }{ \sqrt { \pi n } } \left( 1-\frac { 1 }{ 8n } +\frac { 1 }{ 128n^{ 2 } } +\frac { 5 }{ 1024n^{ 3 } } -\frac { 21 }{ 32768n^{ 4 } } +O(n^{ -5 }) \right)$

where $O(n^{ -5 })$ is the Big O notation. Then we have

$s\sim\sum _{ n=1 }^{ \infty }{ \frac { \frac { 4^{ n } }{ \sqrt { \pi n } } \left( 1-\frac { 1 }{ 8n } +\frac { 1 }{ 128n^{ 2 } } +\frac { 5 }{ 1024n^{ 3 } } -\frac { 21 }{ 32768n^{ 4 } } \right) }{ n\times 4^{ n } } }$

$=\sum _{ n=1 }^{ \infty }{ \frac { 1-\frac { 1 }{ 8n } +\frac { 1 }{ 128n^{ 2 } } +\frac { 5 }{ 1024n^{ 3 } } -\frac { 21 }{ 32768n^{ 4 } } }{ n\sqrt { \pi n } } }$

$=\frac { 1 }{ \sqrt { \pi } } \sum _{ n=1 }^{ \infty }{ \frac { 1-\frac { 1 }{ 8n } +\frac { 1 }{ 128n^{ 2 } } +\frac { 5 }{ 1024n^{ 3 } } -\frac { 21 }{ 32768n^{ 4 } } }{ n\sqrt { n } } }$

$=\frac { 1 }{ \sqrt { \pi } } \left( \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n\sqrt { n } } } -\frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }\sqrt { n } } } +\frac { 1 }{ 128 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 3 }\sqrt { n } } } +\frac { 5 }{ 1024 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 4 }\sqrt { n } } } -\frac { 21 }{ 32768 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 5 }\sqrt { n } } } \right)$

$=\frac { 1 }{ \sqrt { \pi } } \left( \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ \frac { 3 }{ 2 } } } } -\frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ \frac { 5 }{ 2 } } } } +\frac { 1 }{ 128 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ \frac { 7 }{ 2 } } } } +\frac { 5 }{ 1024 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ \frac { 9 }{ 2 } } } } -\frac { 21 }{ 32768 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ \frac { 11 }{ 2 } } } } \right)$

Applying Riemann zeta function, we get

$=\frac { 1 }{ \sqrt { \pi } } \left( \zeta \left( \frac { 3 }{ 2 } \right) -\frac { 1 }{ 8 } \zeta \left( \frac { 5 }{ 2 } \right) +\frac { 1 }{ 128 } \zeta \left( \frac { 7 }{ 2 } \right) +\frac { 5 }{ 1024 } \zeta \left( \frac { 9 }{ 2 } \right) -\frac { 21 }{ 32768 } \zeta \left( \frac { 11 }{ 2 } \right) \right)$

After evaluation, we get approximately $s=1.3867695148$ .

${ e }^{ s }=4.001901$ which is approximately $\boxed{4}$