Evaluate: $\sum_{n=1}^{\infty} {\left(\frac{-100}{729}\right)}^n {3n \choose n}$

Question

The questions asks to evaluate: $$\sum_{n=1}^{\infty} {\left(\frac{-100}{729}\right)}^n {3n \choose n}$$ The answer provided is $-\frac{1}{4}$, but I don't know how to solve it. I am not sure how to approach this series.

Answer 1

A generating function of $\sum_{n=1}^{\infty} {3n \choose n}z^n$ can be derived using Lagrange Inversion and is given as \begin{align*} \sum_{n=0}^{\infty} {3n \choose n}z^n=\frac{2\cos\left(\frac{1}{3}\arcsin\left(\frac{3\sqrt{3}\sqrt{z}}{2}\right)\right)}{\sqrt{4-27z}}\tag{1} \end{align*}

which is shown in this answer.

Here we need (1) evaluated at $z=-\frac{100}{729}$. We obtain \begin{align*} \color{blue}{\sum_{n=1}^{\infty}}\color{blue}{ {3n \choose n}\left(-\frac{100}{729}\right)^n} &=\left.\frac{1}{\sqrt{1-\frac{27}{4}z}}\cos\left(\frac{1}{3}\arcsin\left(\sqrt{\frac{27}{4}z}\right)\right)\right|_{z=-\frac{100}{729}}-1\\ &=\frac{3\sqrt{39}}{26}\cos\left(\frac{1}{3}\arcsin\left(\frac{5}{3\sqrt{3}}i\right)\right)-1\tag{2}\\ &=\frac{3\sqrt{39}}{26}\cos\left(-\frac{i}{3}\ln\left(\sqrt{1+\frac{25}{27}}-\frac{5}{3\sqrt{3}}\right)\right)-1\tag{3}\\ &=\frac{3\sqrt{39}}{26}\cos\left(-\frac{i}{3}\ln\left(\frac{\sqrt{52}-5}{3\sqrt{3}}\right)\right)-1\\ &=\frac{3\sqrt{39}}{52}\left(\left(\frac{\sqrt{52}-5}{3\sqrt{3}}\right)^{\frac{1}{3}}+\left(\frac{\sqrt{52}-5}{3\sqrt{3}}\right)^{-\frac{1}{3}}\right)-1\tag{4}\\ &=\frac{3\sqrt{39}}{52}\left(\frac{1}{\sqrt{3}}\left(\sqrt{52}-5\right)^{\frac{1}{3}}+\frac{\sqrt{3}}{\left(\sqrt{52}-5\right)^{\frac{1}{3}}}\right)-1\tag{5}\\ &=\frac{3\sqrt{13}}{52}\left(\left(\sqrt{52}-5\right)^{\frac{1}{3}}+\left(\sqrt{52}+5\right)^{\frac{1}{3}}\right)-1\tag{6}\\ &=\frac{3}{4}\,\frac{1}{\sqrt{13}}\left(\left(2\sqrt{13}-5\right)^{\frac{1}{3}}+\left(2\sqrt{13}+5\right)^{\frac{1}{3}}\right)-1\\ &=\frac{3}{4}\,\frac{1}{\sqrt{13}}\left(\frac{1}{2}\left(\sqrt{13}-1\right)+\frac{1}{2}\left(\sqrt{13}+1\right)\right)-1\tag{7}\\ &=\frac{3}{4}\,\frac{1}{\sqrt{13}}\sqrt{13}-1\\ &\,\,\color{blue}{=-\frac{1}{4}} \end{align*} and the claim follows.

Comment:

• In (2) we evaluate the function at $z=-\frac{100}{729}$.

• In (3) we use the identity $\arcsin(z)=-i\ln\left(\sqrt{1-z^2}+iz\right),\quad z\ne-1,1$.

• In (4) we use $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.

• In (5) we factor out $\frac{1}{\sqrt{3}}$.

• In (6) we cancel $\sqrt{3}$.

• In (7) we use the identity $2\sqrt{13}\pm5=\left(\frac{1}{2}\left(\sqrt{13}\pm1\right)\right)^3$.