Partial Fraction maybe, or maybe not

Relevant wiki: Fractional Binomial Theorem

Solution from @Akhilesh Prasad, just better presented.

$\begin{aligned} \mathscr I & = 1 + \frac 13 + \frac {1\cdot 3}{3\cdot 6} + \frac {1\cdot 3 \cdot 5}{3\cdot 6 \cdot 9} + \frac {1\cdot 3 \cdot 5 \cdot 7}{3\cdot 6 \cdot 9 \cdot 12} + ... \\ & = \frac {0!!}{3^0 0!} + \frac {1!!}{3^1 1!} + + \frac {3!!}{3^2 2!} + \frac {5!!}{3^3 3!} + \frac {7!!}{3^4 4!} + ... \\ & = \sum_{n=0}^\infty \frac {\color{#3D99F6}{(2n-1)!!}}{3^nn!} \\ & = \sum_{n=0}^\infty \frac {\color{#3D99F6}{(2n)!}}{\color{#3D99F6}{2^nn!}3^nn!} \\ & = \sum_{n=0}^\infty {2n \choose n} \frac 1{2^n3^n} & \small \color{#3D99F6}{\text{As }{-1/2 \choose n} = {2n \choose n} \frac {(-1)^n}{2^{2n}}} \\ & = \sum_{n=0}^\infty {-1/2 \choose n} \frac {(-1)^n2^n}{3^n} \\ & = \left(1-\frac 23\right)^{-\frac 12} = \left(\frac 13\right)^{-\frac 12} = \sqrt{3} \end{aligned}$

$\implies a + b = 2 + 3 = \boxed{5}$ .

The general term of the series can be written as $\displaystyle t_n= \binom {2n} {n} \left(\dfrac{1}{2^n \cdot 3^n} \right)=\dfrac{{(-1)}^n{2}^{n}}{3^n} \displaystyle \binom {-1/2} {n}$ .

Then all we have to find is $\displaystyle \Im = \sum_{n=0}^{\infty}{t_n}$

So, let's begin

$\displaystyle \Im = \sum_{n=0}^{\infty}{t_n}=\sum_{n=0}^{\infty}{ \dfrac{{(-1)}^n{2}^{n}}{3^n} \binom {-1/2} {n}}$

$\Im = {\left(1-\dfrac{2}{3} \right)}^{-1/2}$

From this we can simply conclude $\Im = \sqrt[2]{3}$

Hence, $a=2$ and $b=3$ , so $a+b= \boxed{5}$

Take a look at the following function:

$f(x)=(1+x)^{-\frac{1}{2}} ==> f(0)=1$

$f'(x)=(-\frac{1}{2})*(1+x)^{-\frac{3}{2}} ==> f'(0)=-\frac{1}{2}$

$f''(x)=(-\frac{1}{2})*(-\frac{3}{2})*(1+x)^{-\frac{5}{2}} ==> f''(0)=\frac{1*3}{2^2}$

$f'''(x)=(-\frac{1}{2})*(-\frac{3}{2})*(-\frac{5}{2})(1+x)^{-\frac{7}{2}} ==> f'''(0)=-\frac{1*3*5}{2^3}$

A Maclaurin series is a Taylor series expansion of a function about 0,

$f(x) = f(0) + f'(0)*x + \frac{f''(0)}{2!}*x^2 + \frac{f'''(0)}{3!}*x^3 + ....$

Or in our case

$f(x) = 1 - \frac{1}{2}*x + \frac{1*3}{2^2*2!}*x^2 - \frac{1*3*5}{2^3*3!}*x^3 + ....$

And now for the punchline...

$f(-\frac{2}{3}) = 1 - \frac{1}{2}*(-\frac{2}{3}) + \frac{1*3}{2^2*2!}*(-\frac{2}{3})^2 - \frac{1*3*5}{2^3*3!}*(-\frac{2}{3})^3 + .... =$

$1 + 1/3 + \frac{1*3}{(1*3)*(2*3)} + \frac{1*3*5}{(1*3)*(2*3)*(3*3)} + ....$ =

$1 + 1/3 + \frac{1*3}{3*6} + \frac{1*3*5}{3*6*9} + ....$

And we know that $f(-\frac{2}{3})=(1-\frac{2}{3})^{-\frac{1}{2}} = \frac{1}{3}^{-\frac{1}{2}} = 3^\frac{1}{2} = \sqrt{3}$

$a=2, b=3$ and so the correct answer is

$2+3 = \boxed{5}$