What is the infinite sum $$S = {\frac{2}{5} \\+ \frac{2}{5}\cdot\frac{3}{7} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19}\cdot\frac{17}{23} \\+ \frac{2}{5}\cdot\frac{3}{7}\cdot\frac{5}{11}\cdot\frac{9}{13}\cdot\frac{11}{17}\cdot\frac{15}{19}\cdot\frac{17}{23}\cdot\frac{21}{25} \\+ ...}$$ ?
I tried to figure out the general form of the terms, but this is unfamiliar to me. How should I proceed?
 A: If we group the series in units of pair, the first few pairs are
$$\require{cancel}
\begin{array}{rll}
p_0 :& \frac25 \left(1+\frac37\right) &= \frac25 \frac{10}{7} = \frac47\\
p_1 :& \frac{\color{red}{2}}{\cancel{5}}\frac37\frac{\cancel{5}}{\color{red}{11}}\left(1+\frac{9}{13}\right)
&= \color{red}{\frac{2}{11}}\frac{3}{7}\frac{22}{13} = \frac47\frac{3}{13}\\
p_2 :& \frac{\color{red}{2}}{\cancel{5}}\frac37\frac{\cancel{5}}{\cancel{11}}\frac{9}{13}\frac{\cancel{11}}{\color{red}{17}}
\left(1+\frac{15}{19}\right) &= \color{red}{\frac{2}{17}}\frac{3}{7}\frac{9}{13}\frac{34}{19} = \frac{4}{7}\frac{3}{13}\frac{9}{19}
\end{array}
$$
From this, we see the general pattern for pair $p_n$ is 
$\displaystyle\;\frac{4}{7}\prod_{k=0}^{n-1}\frac{6k+3}{6k+13}$.
The series at hand becomes
$$\sum_{n=0}^\infty p_n 
= \frac47\sum_{n=0}^\infty \prod_{k=0}^{n-1} \frac{6k+3}{6k+13}
= \frac47\sum_{n=0}^\infty \frac{\left(\frac36\right)_n}{\left(\frac{13}{6}\right)_n}
= \frac47 {}_2F_1\left(1,\frac36; \frac{13}{6}; 1 \right)
$$
where $(\alpha)_n = \prod\limits_{k=0}^{n-1}(\alpha+k)$ is the Pochhammer symbol and ${}_2F_1(a,b;c;z)$ is the Hypergeometric function.
WA knows how to evaluate RHS exactly. The result is a surprisely simple number $1$. This suggest we can prove it ourselves. We can simplify above expression using 
Gamma function and Beta function. 
For any $a, b > 0$, we have
$$\frac{(a)_n}{(b)_n} = \frac{\Gamma(b)}{\Gamma(a)}\frac{\Gamma(a+n)}{\Gamma(b+n)}
= \frac{\Gamma(b)}{\Gamma(a)\Gamma(b-a)}\frac{\Gamma(a+n)\Gamma(b-a)}{\Gamma(b+n)}
= \frac{B(a+n,b-a)}{B(a,b-a)}$$
Using the integral representation of beta function
$$B(a,b) = \int_0^1 (1-s)^{a-1} s^{b-1} ds$$
We find
$$\begin{align}
\sum_{n=0}^\infty \frac{(a)_n}{(b)_n} 
&= \frac{1}{B(a,b-a)}\int_0^1 (1-s)^{b-a-1} \sum_{n=0}^\infty s^{a+n-1} ds\\
&= \frac{1}{B(a,b-a)}\int_0^1 (1-s)^{b-a-2} s^{a-1}ds
= \frac{B(a,b-a-1)}{B(a,b-a)}\\
&= \frac{\cancel{\Gamma(a)}\Gamma(b-a-1)}{\Gamma(b-1)}
  \frac{\Gamma(b)}{\cancel{\Gamma(a)}\Gamma(b-a)} = \frac{\Gamma(b-a-1)}{\Gamma(b-a)}\frac{\Gamma(b)}{\Gamma(b-1)}\\
&= \frac{b-1}{b-a-1}
\end{align}
$$
Using this, we find the series at hand equals to
$$\sum_{n=0}^\infty p_n = \frac{4}{7}\left(\frac{\frac{13}{6}-1}{\frac{13}{6}-\frac{3}{6}-1}\right) = 1$$
A: This is more a hint than a full answer, but here is the full series you are looking for with general terms
$$S = \frac{2}{5} + \sum\limits_{n=2}^\infty \left(\frac25 \prod\limits_{k=1}^{n-1} \frac{6k-(-1)^k-1}{6k+(-1)^k+9}\right).$$
