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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.

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    $\begingroup$ en.wikipedia.org/wiki/Mertens%27_theorems makes it clear that Euler didn't have any proof. $\endgroup$ – reuns Jun 18 '20 at 22:41
  • $\begingroup$ Thanks.. kindly look at the edited question. $\endgroup$ – math is fun Jun 18 '20 at 23:08
  • $\begingroup$ @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 $\endgroup$ – Barry Cipra Jun 18 '20 at 23:23
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    $\begingroup$ @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. $\endgroup$ – Barry Cipra Jun 18 '20 at 23:28
  • $\begingroup$ Yeah..I heard it correct $\endgroup$ – math is fun Jun 18 '20 at 23:55
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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 : $$ \sum_{p\leqslant x}\frac{1}{p}\underset{x\rightarrow +\infty}{\sim}\ln\ln x $$

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  • $\begingroup$ 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. $\endgroup$ – reuns Jun 20 '20 at 20:04
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A simple wikipedia goes a long way.

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

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