Doing series is fun!

Relevant wiki: Fractional Binomial Theorem

$\begin{aligned} (1+x)^n & = 1 + \frac {nx}{1!} + \frac {n(n-1)x^2}{2!} + \frac {n(n-1)(n-2)x^3}{3!} + ... & \small \color{#3D99F6} \text{Putting }n = -\frac 12 \\ (1+x)^{-\frac 12} & = 1 - \frac 1{2\cdot 1!}x + \frac {1\cdot 3}{2^2\cdot 2!}x^2 - \frac {1\cdot 3 \cdot 5}{2^3\cdot 3!}x^3 + ... & \small \color{#3D99F6} \text{Putting }x = -\frac 23 \\ \left(1-\frac 23\right)^{-\frac 12} & = 1 + \frac 1{2\cdot 1!}\left(\frac 23 \right) + \frac {1\cdot 3}{2^2\cdot 2!}\left(\frac 23 \right)^2 + \frac {1\cdot 3 \cdot 5}{2^3\cdot 3!}\left(\frac 23 \right)^3 + ... \\ & = 1 + \frac 1{3\cdot 1!} + \frac {1\cdot 3}{3^2\cdot 2!} + \frac {1\cdot 3 \cdot 5}{3^3\cdot 3!} + ... \\ & = 1 + \frac 13 + \frac {1\cdot 3}{3\cdot 6} + \frac {1\cdot 3 \cdot 5}{3\cdot 6 \cdot 9} + ... \end{aligned}$

$\begin{aligned} 1 + \frac 13 + \frac {1\cdot 3}{3\cdot 6} + \frac {1\cdot 3 \cdot 5}{3\cdot 6 \cdot 9} + ... & = \left(1-\frac 23\right)^{-\frac 12} = \left(\frac 13\right)^{-\frac 12} = \boxed{\sqrt 3} \end{aligned}$

Upon observation, the numerator is expressible as:

$\large \frac{(2n)!}{2^n \cdot n!}$

and the denominator as:

$3^n \cdot n!$

which results in the summation:

$\Sigma_{n=0}^{\infty} \frac{1}{3^n \cdot n!} \cdot \frac {(2n)!}{2^n \cdot n!} = \Sigma_{n=0}^{\infty} \binom{2n}{n} (\frac{1}{6})^n = \frac{1}{\sqrt{1 - 4(\frac{1}{6})}} = \frac{1}{\sqrt{1 - \frac{2}{3}}} = \frac{1}{\frac{1}{\sqrt{3}}} =\boxed{\sqrt{3}}.$