<s>[INST] 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!
 [/INST] 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:

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

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