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I don't remember any method to compute the closed from for the following series. $$ \sum_{k=0}^{\infty}\binom{3k}{k} x^k .$$

I tried by putting $\binom{3k}{k}$ in Mathematica for different $k$ and asking for the generating function it deliver a complicated formula which is the following. $$ \frac{2\cos[\frac{1}{3} \sin^{-1}(\frac{\sqrt{3x}}{2})]}{\sqrt{4-27x}} $$

I was wondering if there is any simple form?

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  • $\begingroup$ $$\frac{2 \cos \left(\frac{1}{6} \cos ^{-1}\left(1-\frac{27 x}{2}\right)\right)}{\sqrt{4-27 x}}$$ looks slightly simpler. $\endgroup$ – Claude Leibovici Oct 27 '16 at 10:00
  • $\begingroup$ See oeis.org/A005809 for more information and references $\endgroup$ – Robert Z Oct 27 '16 at 10:01
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That's a power series about $\;x_0=0\;$ and whose sequence of coefficients is

$$a_k=\binom{3k}kx^k=\frac{(3k)!}{k!(2k)!}\implies\;\left|\frac{a_{k+1}}{a_k}\right|=\frac{(3k+3)!}{(k+1)!(2k+2)!}\cdot\frac{k!(2k)!}{(3k)!}|x|=$$

$$=\frac{(3k+1)(3k+2)(3k+3)}{(k+1)(2k+1)(2k+2)}|x|\xrightarrow[k\to\infty]{}\frac{27}{4}|x|$$

and thus the series converges for

$$\frac{27}4|x|<1\iff |x|<\frac4{27}$$

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    $\begingroup$ What? It converges for all $|x|<\frac{4}{27}$. That's what the ratio test tells you. $\endgroup$ – Bob Jones Oct 27 '16 at 19:50
  • $\begingroup$ Stirling's formula tell you there is convergence when $|x| \lt \frac{4}{27}$. $\endgroup$ – Ron Gordon Oct 27 '16 at 20:01
  • $\begingroup$ @BobJones You're right: I forgot completely the $\;x\;$ ...:) :) Editing now, thanks. $\endgroup$ – DonAntonio Oct 27 '16 at 20:03
  • $\begingroup$ @GGT Please do read the edition and correction of my answer. $\endgroup$ – DonAntonio Oct 27 '16 at 20:04
  • $\begingroup$ @RonGordon Thanks, but read above: why would I want to work with Stirling if the ratio test is so easy to apply here...when no mistakes are done...? $\endgroup$ – DonAntonio Oct 27 '16 at 20:05
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Hint: A closed form can be found by means of the Lagrange Inversion Formula.

An answer based upon this method is given at this MSE link.

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