The gamma constant is defined by an equation where the harmonic series is subtracted by the natural logarithm:

$$\gamma = \lim_{n \rightarrow \infty }\left(\sum_{k=1}^n \frac{1}{k} - \ln(n)\right)$$

It is well known that both the harmonic series by itself and the sum of reciprocals of primes are divergent.

Is there a well known function that when subtracted from the sum of reciprocals of primes makes the resultant series convergent?

Is there a function $f(x)$ that makes the following series convergent:

$$\lim_{n \rightarrow \infty }\left(\sum_{p\text{ is a prime }}^n \frac{1}{p} - f(n)\right)$$


4 Answers 4


Yes there is a constant associated with the sum of the reciprocals of the primes. In particular, Mertens showed that $$\sum_{p \text{ prime } \le x} \frac1p - \log\log(x)$$ converges to a constant as $x\to \infty$. This is a result from 1874.

I found the result in a paper:


Mertens' paper is titled: Ein Beitrag zur analytischen Zahlentheorie

  • 1
    $\begingroup$ That is an awesome paper, thanks for sharing it. $\endgroup$
    – Klangen
    Oct 13, 2017 at 10:38

The easiest way to see that $$ -\log n+\sum_{k=1}^{n}\frac{1}{k} $$ is convergent is to write it as $$ -\log\left(1+\frac{1}{n}\right)+\sum_{k=1}^{n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)$$ and check that $\frac{1}{k}-\log\left(1+\frac{1}{k}\right)=O\left(\frac{1}{k^2}\right)$. In the same way, $$ \sum_{p}\left(\frac{1}{p}-\log\left(1+\frac{1}{p}\right)\right) $$ is convergent for sure.

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    $\begingroup$ ... although it does leave you with the problem of estimating $\prod_{p \leq n} (1 + 1/p)$. $\endgroup$
    – user14972
    Jul 4, 2015 at 22:37
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    $\begingroup$ @Hurkyl: true, but $$\prod_{p\leq n}(1+1/p)\approx\prod_{p\leq n}(1-1/p)^{-1}$$ and the RHS is essentially a truncated harmonic series due to Euler's product. $\endgroup$ Jul 4, 2015 at 22:50

A heuristic is that an integer $n$ is prime with "probability" one in $\ln n$, and so we can estimate the sum with its "expected" value:

$$ \sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} \approx \sum_{k=2}^n \frac{1}{k \ln k} \approx \int_2^n \frac{\mathrm{d}x}{x \ln x} \approx \ln \ln n$$

In fact, the Meissel-Mertens constant is given by

$$ M = \lim_{n \to \infty} \left( \sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} - \ln \ln n \right) $$

Another heuristic is that the $n$-th prime is approximately $n \ln n$, so

$$ \sum_{k=1}^n \frac{1}{p_k} \approx \sum_{k=2}^n \frac{1}{k \ln k} \approx \int_2^n \frac{\mathrm{d}x}{x \ln x} \approx \ln \ln n$$

(the lower bound is tweaked to make the sum well-defined; that's okay since the first few terms contribute a fixed value, and so contribute to the total sum in an asymptotically negligible way)

Note that these two heuristics are compatible:

$$ \sum_{k=1}^n \frac{1}{p_k} = \sum_{\substack{p \leq p_n \\ p \text{ prime}}} \frac{1}{p} \approx \ln \ln p_n \approx \ln \ln (n \ln n) = \ln(\ln n + \ln \ln n) \approx \ln \ln n$$

so the difference between "the first $n$ prime reciprocals" and "the prime reciprocals for primes less than $n$" is asymptotically negligible. In fact, we can estimate

$$ \sum_{\substack{n \leq p \leq p_n \\ p \text{ prime}}} \frac{1}{p} \approx \ln \ln p_n - \ln \ln n = \ln \frac{\ln p_n}{\ln n} \approx \ln \frac{\ln n + \ln \ln n}{\ln n} \\ = \ln\left(1 + \frac{\ln \ln n}{\ln n}\right) \approx \frac{\ln \ln n}{\ln n} $$


By Sum of reciprocal prime numbers,

$$\sum_{p \le n}{\frac1{p}} = C + \ln\ln n + O\left(\frac1{\ln n}\right)$$

Therefore $\ln \ln n$ fits the bill.


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