What is the value of the infinite sum of $1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3.5}{3.6.9}+\frac{1.3.5.7}{3.6.9.12}....$

Question

My attempt:

$1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3.5}{3.6.9}+\frac{1.3.5.7}{3.6.9.12}....$

=$1+\frac{1}{6}$.$2\choose 1$+$\frac{1}{6^2}$.$4\choose 2$+$\frac{1}{6^3}$.$8\choose 4$+.......+$\frac{1}{6^{n-1}}$.$2n-2\choose n-1$+...........

But I am perplexed as to what should be my next step?

Options:

(A)$\sqrt{2}$

(B)$\sqrt{3}$

(C)$\sqrt{\dfrac{3}{2}}$

(D)$\sqrt{\dfrac{1}{3}}$

One edit $t_n=\dfrac{1.3.5.7.....(2n-3)}{3^{n-1}(1.2.3....(n-1))}$

$=\dfrac{(2n-2)!}{(6^{n-1}(n-1)!(n-1)!)}$

$=\dfrac{1}{6^{n-1}}$.$2n-2\choose n-1$

Is this going to help me?