The following limit is utilized in Merten’s theorem.

$$ \gamma = \lim\limits_{n\to\infty}\left(-\log\log n - \sum_{p\text{ prime}}^n\log\left(1-\frac{1}{p}\right)\right) $$

I’m interested in the generalization

$$ C_x = \lim\limits_{n\to\infty}\left(x\log\log n - \sum_{p\text{ prime}}^n\log\left(1+\frac{x}{p}\right)\right) $$

For $x\ge 1$. Obviously we can define it in terms of the first equation. In fact for $x=1$ we can say $C_1 = \log\zeta(2) - \gamma$, but for $x\gt 1$ we are left with an unresolved product of primes.

Can we derive a definition of this limit that’s not in terms of primes? Perhaps we can modify a proof of Mertens.

It may be helpful to note that

$$\sum_{k=1}^{\infty} \frac{x^{\Omega(k)}}{k^s} = \prod_{p\text{ prime}}^\infty\left(1-\frac{x}{p^s}\right)^{-1} $$

  • $\begingroup$ Unclear what you expect. $\endgroup$
    – reuns
    Apr 28, 2022 at 19:43
  • $\begingroup$ @reuns I revised my question $\endgroup$
    – tyobrien
    Apr 29, 2022 at 16:38

1 Answer 1


Adding $x$ times the limit that gives $\gamma$ and the one that gives $C_x$ gives

$$C_x+x\gamma=\lim_{n\to\infty}\left(-\sum_{p\leq n}\log\left(1+\frac xp\right)-x\log\left(1-\frac 1p\right)\right)$$

$$=-\log\left(\prod_p \left(1+\frac xp\right)\left(1-\frac 1p\right)^x\right).$$

This expression is easier to search, and this way I found a few threads mentioning it, notably this answer. It doesn't look like it has a known closed form.


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