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 ?

Here is a way that uses the  Gamma function, and that was strangely the first thing to come to mind.  (Warning!:  It is perhaps more complicated than necessary.) 
Let $I$ denote your infinite series. We can write this as $$I=\sum_{n=0}^{\infty}(-1)^{n}\prod_{m=0}^{n}\frac{3m+1}{5m+5}$$ which becomes
$$I=\sum_{n=0}^{\infty}\left((-1)^{n}\frac{3^{n+1}}{5^{n+1}(n+1)!}\prod_{m=0}^{n}\left(m+\frac{1}{3}\right)\right).$$
Now since $$\Gamma\left(\frac{1}{3}\right)\prod_{m=0}^{n}\left(m+\frac{1}{3}\right)=\Gamma\left(n+\frac{4}{3}\right)$$
we have 
$$I=\frac{1}{\Gamma\left(1/3\right)}\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}}{5^{n+1}(n+1)!}\Gamma\left(n+\frac{4}{3}\right).$$
By the definition of $\Gamma(s)$ we have $$\Gamma\left(n+\frac{4}{3}\right)=\int_{0}^{\infty}t^{n+\frac{1}{3}}e^{-t}dt,$$ which implies that $$I=\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-t}\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}t^{n+1}}{5^{n+1}(n+1)!}dt$$ by switching the order of summation and integration. As $$1-\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n+1}t^{n+1}}{5^{n+1}(n+1)!}=\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{n}t^{n}}{5^{n}n!}=e^{-\frac{3t}{5}}$$ it follows that $$I=\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-t}\left(1-e^{-\frac{3t}{5}}\right)dt=1-\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-\frac{8}{5}t}dt$$ Setting $u=\frac{8}{5}t$ we get $$\frac{1}{\Gamma\left(1/3\right)}\int_{0}^{\infty}t^{-\frac{2}{3}}e^{-\frac{8}{5}t}dt=\frac{1}{\Gamma\left(1/3\right)}\left(\frac{5}{8}\right)^{\frac{1}{3}}\int_{0}^{\infty}u^{-\frac{2}{3}}e^{-u}du=\frac{\left(5\right)^{\frac{1}{3}}}{2}$$and hence $$I=1-\frac{\left(5\right)^{\frac{1}{3}}}{2}$$ as desired.
I hope that helps,
