Special series

$Link$ : Central Binomial Coefficient

P.S. I might post my "obtaining process" of generating function of the Catalan sequence later :)

Let's start from $\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2n}x dx =\frac{\left( \begin{matrix} 2n \\ n \end{matrix} \right) }{2^{2n}} \times \frac{\pi}{2}$

And from

$\displaystyle S=1+\frac{1}{5}+\frac{1 \times 3}{5 \times 10}+\frac{1 \times 3 \times 5}{5 \times 10 \times 15}+...=\sum _{ 0 }^{ \infty }{ \frac{\left( \begin{matrix} 2n \\ n \end{matrix} \right) }{10^{n}} }$

Then

$\displaystyle \sum _{ 0 }^{ \infty}{ \int_{0}^{\frac{\pi}{2}} \frac{2^{n}}{5^{n}}\sin^{2n}x dx }= \sum _{ 0 }^{ \infty }{ \frac{\left( \begin{matrix} 2n \\ n \end{matrix} \right) }{10^{n}} } \times \frac{\pi}{2}$

$\displaystyle = \sum _{ 0 }^{ \infty }{ \int_{0}^{\frac{\pi}{2}} \left ( \frac{2}{5}\sin^{2}x \right )^{n} dx }$

$\displaystyle = \int_{0}^{\frac{\pi}{2}}\sum _{ 0 }^{ \infty }{ \left ( \frac{2}{5}\sin^{2}x \right )^{n} } dx$

$\displaystyle = \int_{0}^{\pi/2} \frac{1}{1-\frac{2\sin^{2}x}{5}}dx$

$\displaystyle = \int_{0}^{\pi/2} \frac{1}{1+\frac{3\tan^{2}x}{5}}d\tan x$

$\displaystyle = \sqrt{\frac{5}{3}} \arctan\left ( {\sqrt{\frac{3}{5}}\tan x} \right )$ from 0 to $\pi/2$

Hence

$\displaystyle \sqrt{\frac{5}{3}} \frac{\pi}{2}=S \frac{\pi}{2}$

$\displaystyle S=\sqrt{\frac{5}{3}}$

$\displaystyle \frac{5}{3}=\frac{a}{b}$

$\displaystyle a+b= \boxed{8}$

First, we simplify the series into powers and factorials by removing the $1$ at the start and focusing on $\frac { 1 }{ 5 } +\frac { 1\times 3 }{ 5\times 10 } +\frac { 1\times 3\times 5 }{ 5\times 10\times 15 } +...$

We find that for the denominator,

$5=5\times 1={ 5 }^{ 1 }\times 1!\\ 5\times 10=5\times 5\times 1\times 2={ 5 }^{ 2 }\times 2!\\ 5\times 10\times 15=5\times 5\times 5\times 1\times 2\times 3={ 5 }^{ 3 }\times 3!$

And for the numerator,

$1=1!\div { (2 }^{ 0 }\times 0!)\\ 1\times 3=1\times 2\times 3\div (2)=3!\div (2\times 1)=3!\div { (2 }^{ 1 }\times 1!)\\ 1\times 3\times 5=1\times 2\times 3\times 4\times 5\div (2\times 4)=5!\div (2\times 2\times 1\times 2)=5!\div { (2 }^{ 2 }\times 2!)$

We let $n$ be the number of terms on the numerator (and the denominator).

Therefore we find that

$n=1,\qquad \frac { 1 }{ 5 } =\frac { 1!\div ({ 2 }^{ 0 }\times 0!) }{ { 5 }^{ 1 }\times 1! } =\frac { 1! }{ ({ 2 }^{ 0 }\times 0!)({ 5 }^{ 1 }\times 1!) } \\ n=2,\qquad \frac { 1\times 3 }{ 5\times 10 } =\frac { 3!\div ({ 2 }^{ 1 }\times 1!) }{ { 5 }^{ 2 }\times 2! } =\frac { 3! }{ ({ 2 }^{ 1 }\times 1!)({ 5 }^{ 2 }\times 2!) } \\ n=3,\qquad \frac { 1\times 3\times 5 }{ 5\times 10\times 15 } =\frac { 5!\div ({ 2 }^{ 2 }\times 2!) }{ { 5 }^{ 3 }\times 3! } =\frac { 5! }{ ({ 2 }^{ 2 }\times 2!)({ 5 }^{ 3 }\times 3!) }$

We can now generalise this into

$\frac { (2n-1)! }{ [{ 2 }^{ n-1 }\times (n-1)!]({ 5 }^{ n }\times n!) }$

Hence,

$1+\frac { 1 }{ 5 } +\frac { 1\times 3 }{ 5\times 10 } +\frac { 1\times 3\times 5 }{ 5\times 10\times 15 } +...\\ =1+\sum _{ n=1 }^{ \infty }{ \frac { (2n-1)! }{ [{ 2 }^{ n-1 }\times (n-1)!]({ 5 }^{ n }\times n!) } } \\ =\sqrt { \frac { 5 }{ 3 } } \\ \Rightarrow { \left\{ 1+\sum _{ n=1 }^{ \infty }{ \frac { (2n-1)! }{ [{ 2 }^{ n-1 }\times (n-1)!]({ 5 }^{ n }\times n!) } } \right\} }^{ 2 }=\frac { 5 }{ 3 }$

And because

$\frac { a }{ b } \equiv \frac { 5 }{ 3 }$

So

$\therefore a+b=\boxed { 8 }$