<s>[INST] 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 ?
 [/INST] You use this formula $$(1+x)^{-p/q} = 1 - \frac{\displaystyle\frac{p}{q}}{1!}\cdot x + \frac{ \displaystyle\frac{p}{q} \cdot \Bigl(\frac{p}{q}+1\Bigr)}{2!} x^{2} - \cdots$$
So your series can be rewritten as $$ \frac{1}{3} \times \frac{3}{5} - \frac{\displaystyle\frac{1}{3}\cdot\Bigl(\frac{1}{3} + 1\Bigr)}{2!} \times \Bigl(\frac{3}{5}\Bigr)^{2} + \frac{\displaystyle\frac{1}{3} \cdot \Bigl(\frac{1}{3}+1\Bigr) \cdot \Bigl(\frac{1}{3}+2\Bigr)}{3!} \times \Bigl(\frac{3}{5}\Bigr)^{3} - \cdots$$
So this is nothing but $$\Bigl(1+\frac{3}{5}\Bigr)^{-1/3} -1 =  \frac{1}{3} \times \frac{3}{5} - \frac{\displaystyle\frac{1}{3}\cdot\Bigl(\frac{1}{3} + 1\Bigr)}{2!} \times \Bigl(\frac{3}{5}\Bigr)^{2} + \cdots$$
So our answer will be $$1 - \Bigl(\frac{8}{5}\Bigr)^{-1/3} = 1 -\frac{5^{1/3}}{2}$$
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