Show that $\sum_{n=0}^{\infty} \frac{3^{2n+1}(3n)!}{n!(2n+1)!4^{3n+1}}=1$ I'm trying to show that $$\sum_{n=0}^{\infty} \frac{3^{2n+1}(3n)!}{n!(2n+1)!4^{3n+1}}=1$$
I've tried binomially expanding different expressions to obtain this but I can't seem to find anything, though I don't really have much practice at this sort of thing. 
In particular, I am aware of the identities like 
$$\sum_{n=0}^{\infty} \frac{2^{n+1}(2n)!}{n!(n+1)!3^{2n}}=1,$$ 
but was unable to find anything suitable. 
 A: Define $\, a_n := (3n)!/(n!(2n+1)!)\,$ which is OEIS sequence A001764 and $\, b_n := a_n 3^{2n+1}/4^{3n+1}. \,$
The generating function
$$
 f(x) := \sum_{n=0}^\infty a_n\, x^n = \frac2{\sqrt{3x}}
 \sin\Big(\frac13\, \sin^{-1}\Big(\sqrt{27x/4}\Big)\Big) $$
and thus $$ \sum_{n=0}^\infty b_n\, x^n = \frac34 f\Big(\frac9{64}x\Big) =
  \frac{4}{\sqrt{3x}} \sin\Big(\frac13\, \sin^{-1}
  \Big(\frac{9}{16}\sqrt{3x} \Big)\Big).  $$ 
Set $\, x = 1 \,$ and to prove
$$ \sum_{n=0}^\infty b_n \, =\,
  \frac{4}{\sqrt{3}} \sin\Big(\frac13 \sin^{-1}\Big(\frac{9\sqrt{3}}{16}\Big)\Big) = 1 $$
use the identity
 $\, \sin(3\theta) = 3 \sin(\theta)  - 4\sin(\theta)^3 \,$
where $\, \sin(\theta) = \sqrt{3}/4 \,$ to find
 $\, \sin(3\theta) = 9\sqrt{3}/16. $
In general, for $\, n>1 \,$ we get that $\, f(n^2/(n+1)^3) = (n+1)/n. \,$
Our example is $\, f(9/64) = 4/3.\,$
A: Let's group things a little.
$\sum_{n=0}^{\infty} \frac{3^{2n+1}(3n)!}{n!(2n+1)!4^{3n+1}}
=\dfrac34\sum_{n=0}^{\infty} \frac{3^{2n}(3n)!}{n!(2n+1)!4^{3n}}
=\dfrac34\sum_{n=0}^{\infty} (3^2/4^3)^n\frac{(3n)!}{n!(2n+1)!}
$
so let
$f(x)
=\sum_{n=0}^{\infty} x^n\frac{(3n)!}{n!(2n+1)!}
$.
This looks sort of
like a trisection of series,
but,
being lazy,
I threw it at Wolfy and,
to my great surprise,
 got
$f(x)
=\dfrac{2 \sin(\frac13 \sin^{-1}((3 \sqrt{3x}/2)))}{\sqrt{3x}}
$
which converges when
$|x| < 4/27$.
Putting $x=9/64$,
$\begin{array}\\
f(9/64)
&=\dfrac{2 \sin(\frac13 \sin^{-1}((3 \sqrt{3(9/64)}/2)))}{\sqrt{3(9/64)}}\\
&=\dfrac{2 \sin(\frac13 \sin^{-1}((3 \cdot 3\sqrt{3}/8/2)))}{3\sqrt{3}/8}\\
&=\dfrac{16 \sin(\frac13 \sin^{-1}(9\sqrt{3}/16))}{3\sqrt{3}}\\
&=\dfrac{16 (\sqrt{3}/4)}{3\sqrt{3}}
\qquad\text{again, according to Wolfy}\\
&=\dfrac43\\
\end{array}
$
Muptiplying by $\dfrac34$,
the result is $1$.
As to how I could
prove it by my self,
I don't know.
