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It's well known that the summation over 1/p diverges just as 1/n does. However, in the case of the sum of 1/n, we can establish upper and lower bounds to the sum with the integrals over 1/n and 1/(n-1). Therefore we can say that the sum is asymptotically equal to ln(x). Can we do anything similar for the sum of the reciprocals of the prime numbers?

I suspect there isn't a neat function due to the unpredictable distribution of prime numbers.

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marked as duplicate by Gerry Myerson, Michael Hoppe, Yiorgos S. Smyrlis, mau, Claude Leibovici Feb 13 '14 at 10:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The sum $\Sigma 1/p$ over primes $p\leq n$ is asymptotic to $\ln \ln n$. – Slade Feb 13 '14 at 9:08
This question is a possible duplicate, as pointed out by @GerryMyerson. – Apurv Feb 13 '14 at 9:57

Using $\pi(k)=\frac{k}{\log(k)}\left(1+O\left(\frac1{\log(k)}\right)\right)$, $$ \begin{align} \sum_{p\le n}\frac1p &=\sum_{k=1}^n\frac{\pi(k)-\pi(k-1)}{k}\\ &=\sum_{k=1}^n\frac{\pi(k)}{k}-\sum_{k=0}^{n-1}\frac{\pi(k)}{k+1}\\ &=\frac{\pi(n)}{n}+\sum_{k=1}^{n-1}\frac{\pi(k)}{k(k+1)}\\ &=\frac{\pi(n)}{n}+\sum_{k=1}^{n-1}\frac{\pi(k)}{k^2}-\sum_{k=1}^{n-1}\frac{\pi(k)}{k^2(k+1)}\\ &=\sum_{k=3}^{n-1}\frac{\pi(k)}{k^2}+O(1)\\ &=\sum_{k=3}^{n-1}\left[\frac1{k\log(k)}+O\left(\frac1{k\log(k)^2}\right)\right]+O(1)\\[9pt] &=\log(\log(n))+O(1) \end{align} $$ where we bound the sum by $\int_2^{n-1}\frac{\mathrm{d}x}{x\log(x)}$ and $\int_3^n\frac{\mathrm{d}x}{x\log(x)}$ in the last step.

In fact, $$ \lim_{n\to\infty}\left(\sum_{p\le n}\frac1p-\log(\log(n))\right)=M $$ where $M$ is the Meissel–Mertens constant.

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@BalarkaSen: no bigger than Taylor Series, really, and quite useful. However, for this sum, we could bound by integrals. – robjohn Feb 13 '14 at 16:00
@BalarkaSen: I did use summation by parts at the beginning, or did you mean to sum $\frac1{k\log(k)}$? – robjohn Feb 13 '14 at 16:38

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