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.


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.

  • $\begingroup$ Note that this proof derives the OP's result from an asymptotic formula for $\ell(n)$, but that asymptotic formula itself is nontrivial. $\endgroup$ – Greg Martin Dec 26 '11 at 6:20
  • $\begingroup$ He really should look at Apostol anyway... :) $\endgroup$ – J. M. is a poor mathematician Dec 26 '11 at 6:54

For @clark here is the simple proof i was thinking about : reciprocal of primes serie diverges


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.