Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
As the denominator of the $n$th term $T_n$ is $\displaystyle3\cdot6\cdot9\cdot12\cdots(3n)=3^n \cdot n!$
(Setting the first term to be $T_0=1$)
and the numerator of $n$th term is $\displaystyle1\cdot3\cdot5\cdots(2n-1)$ which is a product of $n$th terms of an Arithmetic Series with common difference $=2,$
we can write $\displaystyle1\cdot3\cdot5\cdots(2n-1)=-\frac12\cdot\left(-\frac12-1\right)\cdots\left(-\frac12-{n+1}\right)\cdot(-2^n)$
which suitably resembles the numerator of Generalized binomial coefficients
$$\implies T_n=\frac{-\frac12\cdot\left(-\frac12-1\right) \cdots\left(-\frac12-{n+1}\right)}{n!}\left(-\frac23\right)^n$$
So, here $\displaystyle z=-\frac23,\alpha=-\frac12$ in $\displaystyle(1+z)^\alpha$
Using Generalized Binomial Expansion, $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$ given the converge holds
Comparing with given Series $\displaystyle nx=\frac13\implies n^2x^2=\cdots\ \ \ \ (1)$
and $\displaystyle\frac{n(n-1)}{2!}x^2=\frac{1\cdot3}{3\cdot6}\ \ \ \ (2)$
Divide $(2)$ by $(1)$ to find $\displaystyle n=-\frac12$ and consequently $\displaystyle x=-\frac23$
Observe that these values satisfy the next two terms, too.
Hence, the sum follows
Consider denominator and numerator separately at first,
$$G_n = 2^n \prod_{m=1}^n m-1/2, \qquad F_n = \frac{1}{3^n n!}$$
Thus we have
$$T_n = \prod_{m=1}^n \frac{m-1/2}{n!} \left(\frac{2}{3}\right)^n \qquad \text{or} \qquad T = \sum_{n=0}^\infty \prod_{m=1}^n \frac{m-1/2}{n!}\left(\frac{2}{3}\right)^n$$ Looking these series elements up we arrive at $T=\sqrt{3}$.
The final form of the series is $$\sum_{n=0}^\infty \frac{\Gamma (n+1/2) }{\sqrt{\pi} n!} \left( \frac{2}{3}\right)^n =\sqrt{3}$$ where $\Gamma(n)$ is the well-known Gamma function.