# Sum of reciprocal prime numbers

How can the following equation be proven?

$$\forall n > 2 : \sum_{p \le n}{\frac1{p}} = C + \ln\ln n + O\left(\frac1{\ln n}\right),$$ where $p$ is a prime number.

It's not homework; I just don't understand from where should I start.

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Apostol gives a proof of this in his book. Here's a more-or-less condensed version:

Letting $[p]$ be an Iverson bracket ($1$ if condition $p$ is true, and $0$ if $p$ is false), we have $\sum\limits_{p \le n}\frac1{p}=\sum\limits_{k \le n}\frac{[k\in\mathbb P]}{k}$ Introduce the function $\ell(n)=\sum\limits_{p \le n}\frac{\log\,p}{p}=\sum\limits_{k \le n}\frac{[k\in\mathbb P]\log\,k}{k}$. Making use of (a special case of) Abel's identity,

$$\sum_{y < n \le x}\frac{a(n)}{\log\,n}=\frac{A(x)}{\log\,x}-\frac{A(y)}{\log\,y}+\int_y^x \frac{A(t)}{t(\log\,t)^2}\mathrm dt$$

where for this case $a(n)=\frac{[n\in\mathbb P]\log\,n}{n}$ and $A(x)=\sum\limits_{k \le x}a(k)$. Taking $y=2$, we have

$$\sum_{p \le n}\frac1{p}=\frac{\ell(n)}{\log\,n}+\int_2^n \frac{\ell(t)}{t(\log\,t)^2}\mathrm dt$$

Since $\ell(n)=\log\,n+O(1)$, we then have

\begin{align*}\sum_{p \le n}\frac1{p}&=1+O\left(\frac1{\log\,n}\right)+\int_2^n \frac1{t\log\,t}\mathrm dt+\int_2^n \frac{\mathfrak{R}(t)}{t(\log\,t)^2}\mathrm dt\\&=1+O\left(\frac1{\log\,n}\right)+\log\log\,n-\log\log\,2+\int_2^n \frac{\mathfrak{R}(t)}{t(\log\,t)^2}\mathrm dt\end{align*}

where $\mathfrak{R}(t)=O(1)$. Since

$$\int_2^n \frac{\mathfrak{R}(t)}{t(\log\,t)^2}\mathrm dt=\int_2^\infty \frac{\mathfrak{R}(t)}{t(\log\,t)^2}\mathrm dt+O\left(\frac1{\log\,n}\right)$$

making the appropriate replacements gives

$$\sum_{p \le n}\frac1{p}=\color{blue}{1-\log\log\,2+\int_2^\infty \frac{\mathfrak{R}(t)}{t(\log\,t)^2}\mathrm dt}+\log\log\,n+O\left(\frac1{\log\,n}\right)$$

where the blue part is the constant term $C$ in the OP.

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Note that this proof derives the OP's result from an asymptotic formula for $\ell(n)$, but that asymptotic formula itself is nontrivial. –  Greg Martin Dec 26 '11 at 6:20
He really should look at Apostol anyway... :) –  Ｊ. Ｍ. Dec 26 '11 at 6:54
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Could you please elaborate a bit? Only a weaker lower bound is proven there. (and if you're only going to post a link, you might as well make it a comment) –  t.b. Dec 25 '11 at 11:16
Oh yes, I am sorry. I need to be careful. At first I thought he was just asking about the lower bound. –  Paul Dec 25 '11 at 14:48