Convergence of $\sum_{n\geq1} \frac{1}{p_n(p_n+1)}$ It is well known that 
$$
\sum_{n\geq1} \frac{1}{n(n+1)}=1,
$$
since 
$$
\sum_{n=1}^{N} \frac{1}{n(n+1)}=1-\frac{1}{N-1}.
$$
But what of 
$$
\sum_{n\geq1} \frac{1}{p_n(p_n+1)},
$$
where $p_n$ denotes the n$^{\rm th}$ prime? More generally, does the sum
$$
S(k)=\sum_{n\geq1} \frac{1}{p_n^k(p_n+1)^k}
$$
converge? If so, to what?

EDIT
Since most comments consist of "it converges using the comparison test", I would like to specify that I am looking for the value of these sums, not a statement about whether they converge or not.
 A: A value with a precision of roughly 683 decimal digits is given in http://oeis.org/A179119 
A: Correct me if wrong:
A proof of the 'trivial':
$a_k= \dfrac{1}{k(k+1)}$, $k=1,2,3,...$
$b_k = \dfrac{1}{p_n(p_n+1)}$ for 
$k=p_n, n=1,2,3...$
$b_k =0$ for $k\not = p_n$, $n=1,2, 3,..$
Then $b_k \le a_k$ , $k \in \mathbb{N}.$
Hence?
A: Let $P(s)=\sum_p\frac{1}{p^s}$ be the prime zeta function.
$$\sum_{p}\frac{1}{p(p+1)} = \sum_{m\geq 2}(-1)^m P(m) = \sum_{m\geq 2}(-1)^m\sum_{k\geq 1}\frac{\mu(k)}{k}\log\zeta(km) $$
allows an efficient numerical evaluation of the LHS, which turns out to be pretty close to $\frac{1}{3}$ ($\approx 0.33023$).
I do not believe there is any "nice" closed form for such series or for the similar Meissel-Mertens constant, but the same approach applies to the series $\sum_{p}\frac{1}{p^s(p+1)^s}$.
A similar technique is to notice that over the interval $\left[0,\frac{1}{5}\right]$ the approximation
$$\frac{x^2}{x+1}\approx \log(1+x^2)-\log(1+x^3)+\frac{3}{2}\log(1+x^4) $$
is extremely accurate, hence
$$ \sum_{p}\frac{1}{p(p+1)}\approx \frac{1}{4}+\sum_{p\geq 5}\left[\log\left(1+\frac{1}{p^2}\right)-\log\left(1+\frac{1}{p^3}\right)+\frac{3}{2}\log\left(1+\frac{1}{p^4}\right)\right] $$
where the RHS can be computed from Euler's product, as
$$\frac{1}{4}-2\log\pi+\log(2^4 3^6 7)+\frac{1}{2}\log(2\cdot 3\cdot 7)-\frac{1}{2}\log(5\cdot 17\cdot 41)-\log(17\cdot 41)-\log\zeta(3)\approx\\ \approx \color{green}{0.330}616\ldots $$
Proving that $\sum_{p\text{ prime}}\frac{1}{p(p+1)}<\frac{1}{3}$ is actually a very nice exercise, thank you for the suggestion.
