To evaluate the sum $\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots$ Right now I am working through archived papers of a math aptitude quiz. For some reason I seem to be haveing a hard time with these series problems. I have managed to write the above series in a compact form but thats as far as I got. 
$$\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots = \sum\limits_{i=0}^\infty (-1)^i\frac{\prod\limits_{j=0}^i (3j+1)}{5^{i+1}(i+1)!}$$
Help please!
 A: Lab Bhattacharjee is right. This is just an instance of the usual binomial series. More precisely,
$$
(1+x)^{-1/3}=1-\frac13x+\frac{1\cdot4}{3^2\cdot 2!}x^2-\frac{1\cdot4\cdot7}{3^3\cdot 3!}x^3+\cdots,
$$
that converges and has the prescribed sum, whenever $-1<x< 1$.
Plugging in $x=3/5$ gives 
$$\frac12\root 3\of5=1-\frac15+\frac{1\cdot4}{5\cdot10}-\frac{1\cdot4\cdot7}{5\cdot10\cdot15}+\cdots.$$ 
This yields $1-\dfrac12\root3\of5$ as the value of the sum.
A: Since:
$$ \prod_{j=0}^{i}\frac{3j+1}{5j+5} = \frac{1}{\left(\frac{5}{3}\right)^{i+1}\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}\cdot B\left(\frac{4}{3}+i,\frac{2}{3}\right) \tag{1}$$
we have:
$$ S=\sum_{i\geq 0}(-1)^i \prod_{j=0}^{i}\frac{3j+1}{5j+5} = \frac{\sqrt{3}}{2\pi}\int_{0}^{1}\sum_{i\geq 0}(-1)^i \left(\frac{3}{5}\right)^{i+1}x^{1/3+i}(1-x)^{-1/3}\,dx \tag{2}$$
hence:
$$ S = \frac{\sqrt{3}}{2\pi}\int_{0}^{1}\frac{3x^{1/3}}{(1-x)^{1/3}(5+3x)}\,dx =\frac{\sqrt{3}}{2\pi}\int_{0}^{+\infty}\frac{3z^{1/3}}{(1+z)(5+8z)}\,dz\tag{3}$$
where in the last step we set $x=\frac{z}{1+z}$. That leads to:
$$ S = \frac{\sqrt{3}}{2\pi}\int_{0}^{+\infty}\frac{9t^3}{(1+t^3)(5+8t^3)}\,dt=\frac{\sqrt{3}}{2\pi}\left(\int_{0}^{+\infty}\frac{3\,dt}{1+t^3}-\int_{0}^{+\infty}\frac{15\,dt}{5+8t^3}\right) \tag{4}$$
and the last integrals can be easily computed through the residue theorem or other techniques, leading to:

$$ S = \color{red}{1-\frac{1}{2}\sqrt[3]{5}}.\tag{5} $$

