Consider the sum $\sum\limits_{p\leq n} \frac{1}{p(p-1)}$, where $p$ are the primes.
This sum certainly converges to some value $<1$ (the sum over all integers is $1$).
Is there a closed form of the limit?
Can it be computed easily (with enough precision)?
Also, is there an asymptotic formula of the sum depending on $n$?
One approach would be the following
$$
\sum\limits_{p\leq n} \frac{1}{p(p-1)} = \sum\limits_{p\leq n} \frac{1}{p-1} - \frac{1}{p}
$$
The latter has an asymptotic expansion of $\log\log(n) + M$ where M is the Meissel-Mertens constant.
I am not sure about the expansion of the former, it is probably along the lines of $\log\log(n) + M'$ with $M'$ the appropriate constant.
I am not sure if there is another closed form for $M'$ (except for $M'=M+\sum\frac{1}{p(p-1)})$, nor if $M,M'$ can be efficiently computed.
It is easy to see that $$ \sum\limits_p {\frac{1}{{p(p - 1)}}} = \sum\limits_{k = 2}^\infty {P(k)} = \sum\limits_{n = 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} , $$ where $P$ is the prime zeta function, $\zeta$ is the Riemann zeta function, and $\mu$ is the Möbius function. Then \begin{align*} \sum\limits_{n = 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} = \;& \sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^{\left\lceil {N/n} \right\rceil + 1} {\log \zeta (nk)} } \right)} \\ & + \sum\limits_{n = N + 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} \\ &+ \sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = \left\lceil {N/n} \right\rceil + 2}^\infty {\log \zeta (nk)} } \right)} \end{align*} for any $N\geq 1$. Now $$ \log \zeta (nk) \le 2 \cdot \frac{1}{{2^{nk} }} $$ for any $n\ge 1$ and $k\ge 2$. Thus for any $n\ge 1$ and $M\ge 1$, $$ \sum\limits_{k = M + 1}^\infty {\log \zeta (nk)} \le 2 \cdot \sum\limits_{k = M + 1}^\infty {\frac{1}{{2^{nk} }}} = 2^{1 - nM} . $$ Therefore $$ \left| {\sum\limits_{n = N + 1}^\infty {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^\infty {\log \zeta (nk)} } \right)} } \right| \le 2\sum\limits_{n = N + 1}^\infty {\frac{1}{{n \cdot 2^n }}} \le 2\int_N^{ + \infty } {\frac{{{\rm d}t}}{{t \cdot 2^t }}} \le \frac{2}{{\log 2}}\frac{1}{{N \cdot 2^N }} $$ and $$ \left| {\sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = \left\lceil {N/n} \right\rceil + 2}^\infty {\log \zeta (nk)} } \right)} } \right| \le 2\sum\limits_{n = 1}^\infty {\frac{1}{{n \cdot 2^{n\left\lceil {N/n} \right\rceil + n} }}} \le 2\frac{1}{{2^N }}\sum\limits_{n = 1}^\infty {\frac{1}{{n \cdot 2^n }}} = \frac{{2\log 2}}{{2^N }} $$ for any $N\ge 1$. Accordingly, $$ \sum\limits_p {\frac{1}{{p(p - 1)}}} = \sum\limits_{n = 1}^N {\frac{{\mu (n)}}{n}\left( {\sum\limits_{k = 2}^{\left\lceil {N/n} \right\rceil + 1} {\log \zeta (nk)} } \right)} + R_N , $$ for any $N\ge 1$, where $$ \left| {R_N } \right| \le \frac{2}{{\log 2}}\frac{1}{{N \cdot 2^N }} + \frac{{2\log 2}}{{2^N }} < \frac{5}{{2^N }}. $$ With $N=3400$, I got \begin{align*}& 0.77315666904979512786436745985594239561874133608318604831100606735670\\& 9028489233397833798758823320818328937814256148184877115780660449046900\\& 2378832590763500601782383946617309470017305607667464101342674273541621\\& 9952460707589642449366317124379893255273683122617198789965614988900346\\& 8789581761777440507530380403930615725533887927380132834996433277138637\\& 7049479823191921992768751966642321281245994223008799785641115700164415\\& 4003845978165611336094017461569027505458605345995912204829815522858359\\& 0259088070578636499788833417902629524916542777136670298457774510084876\\& 5454067301120053261899310892066356658770246502596716068326040486612803\\& 6312902391262059228007144014989942678635902481893404222498689320321229\\& 1181268463226742700101487927499658707519455860084392737464646754241323\\& 4741957037164652252932590566930054867393817757225596885032596671453783\\& 9155845083333132992637494356480167180297031616517608172434964548568691\\& 0734031704796454256219160251853205082189649435393816591198262102370390\\& 29428059110079374440938\ldots \end{align*} for the first $1000$ digits.