Not Wallis Product

$1 + \dfrac{1}{2} \times \left(\dfrac{2}{5}\right)^1 + \dfrac{\dfrac{1}{2}\times \dfrac{3}{2}}{1\times2}\left(\dfrac{2}{5}\right)^2 + \dfrac{\dfrac{1}{2}\times \dfrac{3}{2}\times \dfrac{5}{2}}{1\times2\times 3}\left(\dfrac{2}{5}\right)^3 + ...$

$= 1 + \dfrac{1}{2} \times \left(\dfrac{2}{5}\right)^1 + \dfrac{\dfrac{1}{2}\times\left( 1+ \dfrac{1}{2}\right)}{1\times2}\left(\dfrac{2}{5}\right)^2 + \dfrac{\dfrac{1}{2}\times\left(1 + \dfrac{1}{2}\right)\times \left( \dfrac{1}{2} + 1\right)}{1\times2\times 3}\left(\dfrac{2}{5}\right)^3 + \cdots$

$= \left(1 - \dfrac{2}{5}\right)^{\dfrac{-1}{2}}$

$= \sqrt{ \dfrac{5}{3}}$

Notice that each term is of the following form: $\dfrac{\prod_{k=1}^{n} (2k-1)}{\prod_{k=1}^{n} (5k)}$ $=\dfrac{\prod_{k=1}^{n} (2k-1)*\prod_{k=1}^{n} (2k)}{(n!) 5^n \prod_{k=1}^{n} (2k)}=\dfrac{(2n)!}{(n!)5^n(n!)2^n}=\dfrac{1}{10^n} \binom{2n}{n}$ Hence we need to find $s = \sum_{n=0}^{\infty} \frac{1}{10^n} \binom{2n}{n}$ .

Now since the result is the square root of a rational number, let's find $s^2$ . Using the Cauchy Product (with $\frac{1}{10}$ as the independent variable), we get the following formula: $s^2 = \sum_{n=0}^{\infty} \dfrac{1}{10^n}\sum_{k=0}^{n} \binom{2k}{k} \binom{2(n-k)}{n-k}$ Now it can be shown that for all whole numbers $n$ we have: $\sum_{k=0}^{n}\binom{2k}{k} \binom{2(n-k)}{n-k} = 4^n$ (This can be proved using induction or combinatorics - I will leave it as an exercise.)

Hence we have: $s^2 = \sum_{n=0}^{\infty} \dfrac{1}{10^n}4^n = \sum_{n=0}^{\infty} \left(\dfrac{2}{5}\right)^n = \dfrac{1}{1 - \dfrac{2}{5}} = \dfrac{5}{3}$ Thus $s = \sqrt{\dfrac{5}{3}}$ . Hence $a+b = \boxed{8}$ .

(1+x)^n = 1+nx+n(n-1)x^2 /2 ....find n and x by comparison and u get sqrt(5/3)