A series with prime numbers and fractional parts Considering $p_{n}$ the nth prime number, then compute the limit: 
$$\lim_{n\to\infty} \left\{ \dfrac{1}{p_{1}} + \frac{1}{p_{2}}+\cdots+\frac{1}{p_{n}} \right\} - \{\log{\log n } \}$$
where $\{ x \}$ denotes the fractional part of $x$.  

This is by no means a complete answer but a sketch on how to possibly go about. To get the constant, you need some careful computations.
First get an asymptotic for $ \displaystyle \sum_{n \leq x} \dfrac{\Lambda(n)}{n}$ as $\log(x) - 2 + o(1)$.
To get this asymptotic, you need Stirling's formula and the fact that $\psi(x) = x + o(x)$ i.e. the PNT.
Then relate $ \displaystyle \sum_{\overset{p \leq x}{p- \text{prime}}} \dfrac{\Lambda(p)}{p}$ to $ \displaystyle \sum_{n \leq x} \dfrac{\Lambda(n)}{n}$. Essentially, you can get $$ \displaystyle \sum_{\overset{p \leq x}{p- \text{prime}}} \dfrac{\Lambda(p)}{p} = \displaystyle \sum_{n \leq x} \dfrac{\Lambda(n)}{n} + C + o(1)$$
Getting this constant $C$ is also hard. You can try your luck with partial summation.
Then relate $ \displaystyle \sum_{\overset{p \leq x}{p- \text{prime}}} \dfrac1{p}$ to $\displaystyle \sum_{\overset{p \leq x}{p- \text{prime}}} \dfrac{\Lambda(p)}{p}$ i.e. $\displaystyle \sum_{\overset{p \leq x}{p- \text{prime}}} \dfrac{\log(p)}{p}$ by partial summation to get $$\displaystyle \sum_{\overset{p \leq x}{p- \text{prime}}} \dfrac1{p} = \log(\log(x)) + C_1 + o(1)$$
You might need to invoke PNT here as well. The $C_1$ here obviously depends on he constant $C$ and constant $2$ before.
EDIT Thinking about it there might be a way to avoid PNT to get the constant. 
