Since Mertens showed that $$\sum_{p\le x}\frac{1}{p} = \log\log x + M + O(1/\log x)$$is there a similar equation for $$x\sum_{p\le x}\frac{1}{p-1}$$? ($M$ is the Meissel-Mertens Constant, which is approximately $0.2614972$, and $p$ ranges over the primes less than $x$)

  • $\begingroup$ Sure, $$\sum_p \biggl(\frac{1}{p-1} - \frac{1}{p}\biggr)$$ converges. $\endgroup$ Commented Jul 24, 2017 at 18:50

1 Answer 1


The key observation is that

$$\sum_p \biggl(\frac{1}{p-1} - \frac{1}{p}\biggr)$$

converges. Call its sum $A$. Then we have

\begin{align} \sum_{p \leqslant x} \frac{1}{p-1} &= \sum_{p \leqslant x} \frac{1}{p} + \sum_{p \leqslant x} \biggl(\frac{1}{p-1} - \frac{1}{p}\biggr) \\ &= \sum_{p \leqslant x} \frac{1}{p} + A - \sum_{p > x} \biggl( \frac{1}{p-1} - \frac{1}{p}\biggr). \end{align}

An estimate

$$\sum_{p > x}\biggl(\frac{1}{p-1} - \frac{1}{p}\biggr) \leqslant \sum_{n > x} \biggl(\frac{1}{n-1} - \frac{1}{n}\biggr) = \frac{1}{\lfloor x\rfloor}$$

is immediate, via summation by parts one can obtain the tighter $O\bigl(\frac{1}{x\log x}\bigr)$ estimate for the tail sum, but since the error bound from Mertens' theorem is much larger than the tail sum, this doesn't seem helpful. Overall, we thus have

$$\sum_{p \leqslant x} \frac{1}{p-1} = \log \log x + (M + A) + O\biggl(\frac{1}{\log x}\biggr),$$

only the constant is different from Mertens' theorem.

  • $\begingroup$ Thanks for this step by step justification of your answer. I have two questions before I check this as the correct answer: What is the value of $A$ (or where might I find it)? What do you mean by "tail sum"? $\endgroup$ Commented Jul 24, 2017 at 20:19
  • $\begingroup$ The tail sum is the sum over all $p > x$. Generally, if we have a convergent series $\sum_{n = 1}^{\infty} a_n$, then the "tail sums" of that series are the $\sum_{n = N}^{\infty} a_n$. As for the value of $A$, I don't think there's a closed form. It may well be a named constant, but I don't know the name if it has one. $\endgroup$ Commented Jul 24, 2017 at 20:29
  • $\begingroup$ The value is approximately $0.7731566689$. $\endgroup$ Commented Jul 24, 2017 at 20:37
  • $\begingroup$ Thanks so much for this clarification. This all makes sense now. Where was it that you found the value of the constant, and from which source did you find that the sum $$\sum_p \biggl(\frac{1}{p-1} - \frac{1}{p}\biggr)$$ converges? I'm wondering about this since I'd like to build a little bank of references to drawn on in the future. Thanks again for the great answer! $\endgroup$ Commented Jul 24, 2017 at 23:24
  • $\begingroup$ I didn't find the value of the constant, I wrote a small programme to (approximately) compute it. Since $\frac{1}{p-1} - \frac{1}{p} = \frac{1}{p(p-1)} \approx \frac{1}{p^2}$, the convergence of that series follows immediately from the convergence of $\sum \frac{1}{n^a}$ for $a > 1$. $\endgroup$ Commented Jul 25, 2017 at 11:10

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