Show that $\sum_{n=0}^{\infty} \frac{3^{2n+1}(3n)!}{n!(2n+1)!4^{3n+1}}=1$

Question

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.

Answer 1

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.

Answer 2

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 $\, f(9/64) = 4/3.\,$