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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 $x$.

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Do you know that the series diverges like $\log \log n$, for starters? That is $$\sum_{n=1}^n \frac 1 {p_n} \sim \log \log n$$ –  Pedro Tamaroff Jun 15 '12 at 22:48
    
@PeterTamaroff That doesn't really help though. It doesn't even show existence of the limit, e.g. $2^n$ diverges like $2^n+n$. –  Alex Becker Jun 15 '12 at 22:50
    
The limit does exist. However, to get an estimate of the limit you need to prove prime number theorem i.e. for this case we need to prove that $\psi(x) \sim x$. –  user17762 Jun 15 '12 at 23:05
    
@AlexBecker I'm talking about $o(1)$ here. –  Pedro Tamaroff Jun 15 '12 at 23:06

2 Answers 2

In Wikipedia under Third it is stated that without the fractional part signs the limit is the Meissel–Mertens constant, about $0.261497$. There must be many $n$ for which the first term is greater than some natural and the second term is less than the same and many other $n$ where they are between the same pair of naturals. In the first case the value will be about $0.261497-1$, in the second it will be about $0.261497$, so the limit will not exist.

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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.

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It's worth pointing out that the asymptotic for $\sum 1/p$ was proved by Mertens long before PNT, including (miraculously) the constant. –  Erick Wong Jun 16 '12 at 6:31
    
@ErickWong So the PNT is not needed in finding the constant? –  user17762 Jun 16 '12 at 6:54
    
Correct, although if memory serves it relies on some miraculous identities and/or surprising cancellations. Oh, this paper by Mark Villarino discusses Mertens's original proof, very nice! –  Erick Wong Jun 16 '12 at 7:16

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