How can I sum the infinite series  $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$ How can I find the sum of the infinite series 
$$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$
My attempt at a solution - I saw that I could rewrite it as 
$$\frac{1}{5}\left(1 - \frac{4}{10} \left( 1 - \frac{7}{15} \left(\cdots \left(1 - \frac{3k - 2}{5k}\left( \cdots \right)\right)\right.\right.\right.$$ 
and that $\frac{3k - 2}{5k} \to \frac{3}{5}$ as $k$ grows larger. Using this I thought it might converge to $\frac{1}{8}$, but I was wrong, the initial terms deviate significantly from $\frac{3}{5}$.
According to Wolfram Alpha it converges to $1-\frac{\sqrt[3]{5}}{2}$. How can I get that ?

\begin{align*}(-1)^{n-1}\frac{1\cdot 4 \dots (3n-2)}{5\cdot 10 \dots 5n}&= 
(-1)^{n-1}\frac{3^n}{5^n}(-1)^n\frac{(-\frac{1}{3})\cdot (-\frac{4}{3}) \dots (-\frac{3n-2}{3})}{1\cdot 2 \dots n}\\ &= -(3/5)^n\binom{-1/3}{n}\end{align*}
Therefore, you can obtain
$$\sum_{n=1}^{\infty} -(3/5)^n\binom{-1/3}{n} = 1 - \sum_{n=0}^{\infty} (3/5)^n\binom{-1/3}{n} = 1- (1+\frac35)^{-1/3} = 1- \sqrt[3]{5/8} =1-\frac{\sqrt[3]5}{2}$$
