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? $ 
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$

I can see that 3 cancels out after 1/3, but what next? I can't go further.
 A: \begin{align*}
  1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+
  \frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+\ldots
  &=\sum_{n=0}^{\infty}
    \frac{(2n-1)!!}{3^{n} n!} \\
  &=\sum_{n=0}^{\infty}
    \frac{(-\frac{1}{2})(-\frac{3}{2})\ldots (-\frac{2n-1}{2})}
         {3^{n} n!} (-2)^{n}   \\
  &=\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}
  \frac{ \left( -\frac{2}{3} \right)^{n}}
       {n!} \\
  &= \left[ 1+\left( -\frac{2}{3} \right) \right]^{-\frac{1}{2}} \\
  &= \sqrt{3}
\end{align*}
A: HINT:
Look at the Taylor series of $f(x)=(1-x)^{-1/2}$ evaluated at $x=1/3$. 


Note that the general terms of the series can be written $$a_n=\frac{(2n-3)!!}{3^{n-1}(n-1)!}$$The series can be written as $$1+\sum_{n=1}^\infty \frac{(2n-1)!!}{3^n\,n!}=\sum_{n=0}^\infty \frac{(2n)!}{4^n\,(n!)^2}\left(\frac23\right)^n$$which we recognize as the series for $(1-x)^{-1/2}$ evaluated at $x=2/3$.  Therefore, the answer is (c) $\sqrt{3}$

A: From the OGF of Catalan numbers we have that:
$$ \sum_{n\geq 0}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} $$
where the radius of convergence of the LHS is $\frac{1}{4}$ since $\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{4n+2}{n+1}$.
By evaluating the previous identity at $x=\frac{1}{6}$ it follows that:
$$ \color{red}{\sqrt{3}} = 1+\sum_{n\geq 1}\frac{(2n)!}{n! n! 6^n} = 1+\sum_{n\geq 1}\frac{(2n-1)!!}{3^n n!} $$
where the RHS is exactly our sum.
A: 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 $\displaystyle(1+z)^\alpha$
A: 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  
A: 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 $T=\sqrt{3}$.
EDIT
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.
A: Here is another approach.  The answer equals $f(1)$ where
$$\begin{align}f(x)&=1+\frac13x+\frac{1\cdot 3}{3\cdot 6}x^2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^3+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}x^4+\cdots\\
\frac{df}{dx}&=\frac13\cdot 1+\frac{1\cdot 3}{3\cdot 6}x\cdot 2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^2\cdot 3+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}x^3\cdot 4+\cdots\\
3\frac{df}{dx}&=1+\frac13x\cdot 3+\frac{1\cdot 3}{3\cdot 6}x^2\cdot 5+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^3\cdot 7+\cdots\\
&=f(x)+2x\left(\frac13\cdot 1+\frac{1\cdot 3}{3\cdot 6}x\cdot 2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^2\cdot 3+\cdots\right)\\
3\frac{df}{dx}&=f(x)+2x\frac{df}{dx}\end{align}$$
Now solve the initial value problem with $f(0)=1$.
$$\begin{align}(3-2x)\frac{df}{dx}&=f \\
\frac{df}{f}&=\frac{dx}{3-2x} \\
\ln f&=\ln [C(3-2x)^{-1/2}]\\
f(x)&=C(3-2x)^{-1/2}\\
f(0)&=1=C\cdot 3^{-1/2} \\
C&=3^{1/2} \\
f(x)&=3^{1/2}(3-2x)^{-1/2}\\
f(1)&=3^{1/2}.\end{align}$$
