Limits and binomial theorem

$\begin{aligned} S & = 1 + \sum_{r=1}^\infty \frac 1{3^r r!}\prod_{i=1}^r (2i-1) \\ & = 1 + \sum_{r=1}^\infty \frac {(2r-1)!!}{3^r r!} \\ & = 1 + \sum_{r=1}^\infty \frac {(2r)!}{3^r \cdot 2^r (r!)^2} \\ & = 1 + \sum_{r=1}^\infty \binom {2r}r \cdot \frac 1{6^r} & \small \blue{\text{Since }\sum_\red{n=0}^\infty \binom {2n}n x^n = \frac 1{\sqrt{1-4x}}} \\ & = \sum_\red{r=0}^\infty \binom {2r}r \cdot \frac 1{6^r} \\ & = \frac 1{\sqrt{1-\frac 23}} = \sqrt 3 \approx \boxed{1.73} \end{aligned}$

given series can be written as

$\begin{aligned}S = \sum_{r=0}^{\infty} \binom{2r}{r} \frac{1}{6^{r}} \end{aligned}$

which is equal to the coeffcient of $x^{0}$ in the expansion of

$\begin{aligned}f(x) = \sum_{k=0}^{\infty}\left(\frac{1}{x} +\frac{x}{6}\right)^{2k} \end{aligned}$

and for $\begin{aligned}0<|\left(\frac{1}{x} +\frac{x}{6}\right)|<1\end{aligned}$

$\begin{aligned}f(x)=\frac{1}{1-\left(\frac{6+x^{2}}{6x}\right)^{2}}\end{aligned}$

$\begin{aligned}f(x)=\frac{-36 x^{2}}{x^{4}-24x^{2}+36}\end{aligned}$

$\begin{aligned}f(x)=\frac{36 x^{2}}{(x^{2}-\alpha)(\beta-x^{2})}\end{aligned}$

$\begin{aligned}f(x)=\frac{36 x^{2}}{\alpha-\beta}\left(\frac{1}{x^{2}-\alpha}+\frac{1}{\beta-x^{2}}\right)\end{aligned}$

$\begin{aligned}f(x)=\frac{36 }{\alpha-\beta}\left(\frac{1}{1-\frac{\alpha}{x^{2}}}+\frac{x^{2}}{\beta(1-\frac{x^{2}}{\beta})}\right)\end{aligned}$

$\begin{aligned}f(x)=\frac{36 }{\alpha-\beta}\sum_{k=0}^{\infty}\left(\left(\frac{\alpha}{x^{2}}\right)^{k}+\left(\frac{x^{2}}{\beta}\right)^{k+1}\right)\end{aligned}$

coffcient of $x^{0}$ is $\begin{aligned}\frac{36}{\alpha- \beta}\end{aligned}$ and $\alpha- \beta =12\sqrt{3}$

so answer is $\sqrt{3}$