Evaluate$\sum_{n=0}^\infty\binom{3n}{n}x^n$

Question

The question is:

Evaluate $$\sum_{n=0}^\infty\binom{3n}{n}x^n$$

After applying a few numbers as $x$ in Wolfram Alpha, I guess that the answer is probably: $$2\sqrt{\frac1{4-27x}}\cos\left( \frac13\sin^{-1}\frac{3\sqrt{3x}}{2} \right)$$ that I can never prove. (Interestingly the above becomes simply $2\cos\frac{\pi}9$ when $x=\frac19$.)

(*) To give you the background, the motivation that led me to this question is the Algebra problem #$10$ in the Harvard-MIT Math Test in Feb. 2008, that concludes to: $$\sum_{n=0}^\infty\binom{2n}{n}x^n=\frac1{\sqrt{1-4x}}$$ And then I thought about what if it was $3n$ instead of $2n$.

Answer 1

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Lets $\ds{\,\mrm{f}\pars{x} = \sum_{n = 0}^{\infty}{3n \choose n}x^{n}. \qquad\mrm{f}\pars{0} = 1}$.

Then, \begin{align} \mrm{f}'\pars{x} & = \sum_{n = 1}^{\infty}{3n \choose n}nx^{n - 1} = \sum_{n = 0}^{\infty}{3n + 3 \choose n + 1}\pars{n + 1}x^{n} = \sum_{n = 0}^{\infty}{\pars{3n + 3}! \over n!\pars{2n + 2}!}\,x^{n} \\[5mm] & = \sum_{n = 0}^{\infty}{\pars{3n + 3}\pars{3n + 2}\pars{3n + 1}\pars{3n}! \over n!\pars{2n + 2}\pars{2n + 1}\pars{2n}!}\,x^{n} = {3 \over 2}\sum_{n = 0}^{\infty}{\pars{3n + 2}\pars{3n + 1} \over 2n + 1} {3n \choose n}\,x^{n} \\[5mm] & = {27 \over 8}\ \underbrace{\sum_{n = 0}^{\infty}{3n \choose n}\,x^{n}}_{\ds{\mrm{f}\pars{x}}}\ +\ {27 \over 4}\ \underbrace{\sum_{n = 0}^{\infty}{3n \choose n}\,n\,x^{n}} _{\ds{x\,\mrm{f}'\pars{x}}}\ -\ {3 \over 8}\sum_{n = 0}^{\infty}{1 \over 2n + 1}{3n \choose n}\,x^{n} \label{1}\tag{1} \end{align}

where $\ds{\mrm{f}'\pars{0} = 3}$.

Note that \begin{align} \sum_{n = 0}^{\infty}{1 \over 2n + 1}{3n \choose n}\,x^{n} & = \sum_{n = 0}^{\infty}{3n \choose n}\,x^{n}\int_{0}^{1}t^{2n}\,\dd t = \int_{0}^{1}\sum_{n = 0}^{\infty}{3n \choose n}\,\pars{xt^{2}}^{n}\,\dd t = \int_{0}^{1}\mrm{f}\pars{xt^{2}}\,\dd t \\[5mm] & = {1 \over 2}\,x^{-1/2}\int_{0}^{x}{\mrm{f}\pars{t} \over t^{1/2}}\,\dd t \end{align}

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Expression \eqref{1} is reduced to \begin{align} \pars{16x^{1/2} - 108x^{3/2}}\,\mrm{f}'\pars{x} & = 54x^{1/2}\,\mrm{f}\pars{x} -3\int_{0}^{x}{\mrm{f}\pars{t} \over t^{1/2}}\,\dd t \end{align} Moreover, \begin{align} &\pars{8x^{-1/2} - 162x^{1/2}}\,\mrm{f}'\pars{x} + \pars{16x^{1/2} - 108x^{3/2}}\,\mrm{f}''\pars{x} \\[5mm] = &\ 27x^{-1/2}\,\mrm{f}\pars{x} + 54x^{1/2}\,\mrm{f}'\pars{x} - 3\,\mrm{f}\pars{x}x^{-1/2} \end{align} and \begin{equation}\bbx{\ds{ \pars{16x - 108x^{2}}\,\mrm{f}''\pars{x} + \pars{8 - 216x}\,\mrm{f}'\pars{x} - 24\,\mrm{f}\pars{x} = 0\,,\qquad \left\{\begin{array}{rcl} \ds{\mrm{f}\pars{0}} & \ds{=} & \ds{1} \\[2mm] \ds{\mrm{f'}\pars{0}} & \ds{=} & \ds{3} \end{array}\right.}}\label{2}\tag{2} \end{equation} The solution of differential equation \eqref{2} is given by: $$\bbox[#ffe,25px,border:1px dotted navy]{\ds{% {\root{3} \over \root{4 - 27x}}\bracks{% \cos\pars{{1 \over 3}\,\mrm{arccsc}\pars{2 \over \root{4 - 27x}}} + \sin\pars{{1 \over 3}\,\mrm{arccsc}\pars{2 \over \root{4 - 27x}}}}}} $$ An equivalent expresion is $$\bbox[#ffe,25px,border:1px dotted navy]{\ds{% {\root{6} \over \root{4 - 27x}} \cos\pars{{1 \over 3}\,\mrm{arccsc}\pars{% 2 \over \root{4 - 27x}} - {\pi \over 4}}}} $$

Answer 2

We proceed with complex analysis. Assume that the series converges.

$$\binom{3n}{n} x^n =\frac{1}{2\pi i}\int_{|z|=1} x^n\frac{(z+1)^{3n}}{z^{n+1}}dz =\frac{1}{2\pi i}\int_{|z|=1} \left(\frac{x(z+1)^3}{z}\right)^n\frac{1}{z} dz$$

Thus, $$\sum_n \binom{3n}nx^n= \frac{1}{2\pi i}\int_{|z|=1} \frac{1}{z}\sum_n\left(\frac{x(z+1)^3}{z}\right)^n dz \\= \frac{1}{2\pi i}\int_{|z|=1}-\frac{1}{x z^3 + 3 x z^2 + 3 x z + x - z} dz $$

Now, we can use the cubic formula and find where the integrand has singularities, then use the residue theorem in order to find the value of the contour integral, and thus the value of the sum. Unfortunately, these computations are going to be very tedious, though doable, and the result should be the same as what you have previously gotten.

Notice that in the case of $\binom{2n}{n},$ the above contour integral would only have a quadratic function in its denominator, and thus it would have been trivial to solve that integral, which explains why the formula for $\binom{3n}{n}$ is much more complicated than that of $\binom{2n}{n}$

Answer 3

This is too long for a comment.

Considering $$f_k=\sum_{n=0}^\infty\binom{kn}{n}x^n$$ where $k$ is a positive integer,we could notice nice patterns using the generalized hypergeometric function $$f_4=\, _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{1}{3},\frac{2}{3};\frac{4^4 }{3^3}x\right)$$ $$f_5=\, _4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{4},\frac{2}{ 4},\frac{3}{4};\frac{5^5 }{4^4}x\right)$$ $$f_6=\, _5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{1}{ 5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{6^6 }{5^5}x\right)$$ As you already noticed, this only simplifies for $k=2$ and $k=3$.

Answer 4

By Lagrange's inversion theorem $$ f(z) = \sum_{k\geq 0}\binom{3k}{k}\frac{z^{2k+1}}{2k+1} $$ is the inverse function of $x-x^3$, and you are essentially dealing with $f'(z)$.
Through implicit differentiation it is not difficult to spot the connection between your series and the (trigonometric) solutions of a cubic polynomial. See also Bring radical.