Is there a function that can be subtracted from the sum of reciprocals of primes to make the series convergent

Question

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:

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

Answer 1

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:

EULER’S CONSTANT: EULER’S WORK AND MODERN DEVELOPMENTS - By JEFFREY C. LAGARIAS

Mertens' paper is titled: Ein Beitrag zur analytischen Zahlentheorie

Answer 2

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.

Answer 3

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

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

Answer 4

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.