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.

Where did you find this? I "proved" it with Wolfy's help, but two of the steps are magic. See my answer. — marty cohen, Jul 17 '18 at 02:19
simplifize $\displaystyle \sum {_{3n}C_n}\frac{3^{2n+1}}{(2n+1)4^{3n+1}}$ — Takahiro Waki, Jul 17 '18 at 02:58
This sum arises combinatorially from the generating function for the Fuss-Catalan Numbers. The desired value can be obtatined by calculating $\frac{3}{4} C^{(3)}(\frac{9}{64})$, using the associated cubic equation provided in the linked answer. — Paul LeVan, Jul 17 '18 at 03:01
This series is seen in problem 45 (ii) of Advanced Problems in Mathematics. It is a result from applying Lagrange inversion theorem to find a root of the equation $x-x^3/4=3/4$. — zytsang, Aug 28 '21 at 19:12

Somos · Answer 1 · 2018-07-17T11:49:08.610

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

One is to look in OEIS which has the g.f. given. Another is to use a CAS such as Maple or Mathematica. There are other ways but they are harder and require much more knowledge about enumberative combinatorics, for example. See the almost 100 references and links in the OEIS entry — Somos, Jul 17 '18 at 22:38

score 0 · Answer 2 · answered Jul 17 '18 at 02:17

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.

Amazing ! I directly gave a CAS $$f(x)=\sum_{n=0}^{\infty} \frac{3^{2n+1}(3n)!}{n!(2n+1)!4^{3n+1}}x^n$$ and received your answer. — Claude Leibovici, Jul 17 '18 at 02:50

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

2 Answers2