Divergence of the Sum of the Reciprocal of the Primes

Question

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 $x$)

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.