# $\sum_{p\leq x} \frac{1}{p}$ ~$\log \log x$ as $x\to \infty$.

I have heard that, Euler proved $$\sum_{p\leq x} \frac{1}{p}$$ ~$$\log \log x$$ as $$x\to \infty$$.

Can anyone please refer me the paper or any link where I can find the proof?

I know there are standard alternative proofs available. But I am really curious to know Euler's approach to this problem.

Edit: I might have missed it. I think it was $$\sum_{p\leq x} \frac{1}{p}$$ ~$$\log x$$ as $$x\to \infty$$ which Euler proved. Can I get that proof?

Any help would be appreciated. Thanks in advance.

• en.wikipedia.org/wiki/Mertens%27_theorems makes it clear that Euler didn't have any proof. – reuns Jun 18 '20 at 22:41
• Thanks.. kindly look at the edited question. – math is fun Jun 18 '20 at 23:08
• @reuns, but Euler did seem to know the theorem, if not a proof: See the top of page 3 at arxiv.org/pdf/math/0504289.pdf – Barry Cipra Jun 18 '20 at 23:23
• @mathisfun, your edit makes little sense. The sum is asymptotic to $\ln\ln x$, not $\ln x$. It's unlikely Euler would have guessed wrong, much less proved the wrong result. – Barry Cipra Jun 18 '20 at 23:28
• Yeah..I heard it correct – math is fun Jun 18 '20 at 23:55

Let $$g(x):=\sum_{p\leqslant x}\frac{\ln p}{p}$$ for $$x\geqslant 2$$, where $$p$$ denotes a prime number. First, we show the lemma $$g(x)=\ln x+\mathcal{O}(1)$$. Let $$n\geqslant 2$$, then $$v_p(n!)=\sum_{k=1}^{+\infty}\left\lfloor\frac{n}{p^k}\right\rfloor$$ Since $$\frac{n}{p^k}-1\leqslant\left\lfloor\frac{n}{p^k}\right\rfloor\leqslant\frac{n}{p^k}$$ we have $$\frac{n}{p}-1\leqslant v_p(n!)\leqslant\frac{n}{p-1}$$, and since $$\ln(n!)=\sum_{p\leqslant n}v_p(n!)\ln p$$, we have $$\ln(n!)\geqslant ng(n)-\sum_{p\leqslant n}\ln p\geqslant ng(n)-\pi(n)\ln n=n\Big(g(n)+\mathcal{O}(1)\Big)$$ because $$\pi(n)=\mathcal{O}\left(\frac{n}{\ln n}\right)$$ (it follows from $$\prod_{n< p\leqslant 2n}p\leqslant\binom{2n}{n}$$). We also have $$\ln(n!)\leqslant n\sum_{k\leqslant n}\frac{\ln p}{p-1}=ng(n)+n\sum_{p\leqslant n}\frac{\ln p}{p(p-1)}=n\Big(g(n)+\mathcal{O}(1)\Big)$$ because $$\sum_{p}\frac{\ln p}{p(p-1)}\leqslant\sum_{k\geqslant 2}\frac{\ln k}{k(k-1)}<+\infty$$ By Stirling's formula we have $$\ln(n!)=n\ln n+\mathcal{O}(n)$$ and thus $$g(n)=\ln n+\mathcal{O}(1)$$. For $$x\geqslant 2$$ we have $$g(x)=\ln\lfloor x\rfloor+\mathcal{O}(1)=\ln x+\mathcal{O}(1)$$.
Now \begin{aligned} \sum_{p\leqslant x}\frac{1}{p}&=\sum_{n\leqslant x}\frac{g(n)-g(n-1)}{\ln n} \\ &=\sum_{n\leqslant x}\frac{g(n)}{\ln n}-\sum_{n\leqslant x-1}\frac{g(n)}{\ln(n+1)} \\ &=\sum_{n\leqslant x}g(n)\left(\frac{1}{\ln n}-\frac{1}{\ln(n+1)}\right)+\mathcal{O}\left(\frac{g(x)}{\ln x}\right) \\ &=\sum_{n\leqslant x}\left(1-\frac{\ln n}{\ln(n+1)}\right)+\mathcal{O}(1) \end{aligned} because of the lemma. But $$1-\frac{\ln n}{\ln(n+1)}\underset{n\rightarrow +\infty}{\sim}\frac{1}{n\ln n}$$ and since $$\sum_{n\geqslant 2}\frac{1}{n\ln n}$$ diverges, we have $$\sum_{n\leqslant x}\left(1-\frac{\ln n}{\ln(n+1)}\right)\underset{x\rightarrow +\infty}{\sim}\sum_{n\leqslant x}\frac{1}{n\ln n}\underset{x\rightarrow +\infty}{\sim}\int_2^x\frac{dt}{t\ln t}\underset{x\rightarrow +\infty}{\sim}\ln\ln x$$ This leads to the desired result : $$\sum_{p\leqslant x}\frac{1}{p}\underset{x\rightarrow +\infty}{\sim}\ln\ln x$$
• For the readability : starting with $$\log n\sim \frac{\log n! }{n}= \sum_{p \le n} \log p \sum_{k\ge 1} \frac{[n/p^k]}{n} =\sum_{p \le n} \log p \frac{1+o(1)}p=(1+o(1))\sum_{p \le n}\frac{\log p}p$$ is a very good idea, we are mostly done from there as the result follows by partial summation. – reuns Jun 20 '20 at 20:04