Convergence of sum related to $\sum_k 1/p_k$ Motivation (skip if desired): We can show that $$S =\frac{1}{2}+ \frac{1}{2\cdot 3}+ \frac{(2)}{2\cdot (3)\cdot5}+\frac{(2\cdot 4)}{2\cdot(3\cdot5)\cdot 7}+\frac{(2\cdot4\cdot6)}{2\cdot(3\cdot5\cdot7)\cdot11}+...$$etc. converges. 
We know
$$\frac{1}{2}\sum_{k=1}\frac{1}{p_k} = \infty$$ and 
$$ \sum_{n=1}^{\infty} \frac{1}{n} = \infty$$
and that the sum of factors (in parentheses above)
$$ \frac{2}{3} + \frac{2\cdot 4}{3\cdot 5} + \frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}+\frac{2\cdot 4\cdot6\cdot 10}{3\cdot 5\cdot 7\cdot 11}+... = \infty.$$
Question: Let $a(3) = \frac{2}{3}, a(4) = \frac{2\cdot 4}{3\cdot 5},$ etc. Does $$T = \frac{1}{2}+ \frac{1}{2\cdot 3}+\frac{2}{2\cdot3\cdot 5}+\frac{2\cdot 4}{2\cdot3\cdot5\cdot6}+\frac{2\cdot 4\cdot6}{2\cdot3\cdot5\cdot7\cdot7}+~ ...~\text{etc.}$$
$$= \frac{1}{2}+ \frac{1}{2\cdot 3}+\frac{1}{2}\sum_{k=3}\frac{a(k)}{k+2} $$ 
converge?
Note: $a(n) = \prod_{k=3}^n \frac{p_{k-1}-1}{p_{k-1}}.$ 

As the commenters have suggested: Mertens proved the asymptotic formula
$$
\prod_{p\le x} \bigg( 1-\frac1p \bigg) \sim \frac{e^{-\gamma}}{\log x}
$$
(where $\gamma$ is Euler's constant). Therefore
$$
\prod_{k=3}^{n} \frac{p_{k-1}-1}{p_{k-1}} = 2 \prod_{p\le p_{n-1}} \bigg( 1-\frac1p \bigg) \sim \frac{2e^{-\gamma}}{\log p_{n-1}} \sim \frac{2e^{-\gamma}}{\log n}.
$$
Therefore
$$
\frac{a(k)}{k+2} \sim \frac{2e^{-\gamma}}{k \log k},
$$
and so $\sum_k \frac{a(k)}{k+2}$ diverges by the limit comparison test.
