While working on an exercise on the Dirichlet series for various sequences I had some trouble finding it for the sequence $\{ ln(n) \}_1^{\infty}$ because I forgot to use my brain. Due to this I ended up doing something completely different and, as a result, wounded up with the identity:

$$ \sum_{n=1}^{\infty} \frac{\ln(n)}{n^s} = \zeta(s) \cdot \sum_{p} \frac{\ln(p)}{p^s-1} $$

instead. Afterwards, I wanted to make sure it was correct (and maybe even if there were more ways to simplify it or other fun identities like this), but even after searching in many different ways I was unable to find it mentioned any other place. After some testing I can say that it has a higher rate of convergence compared to the original expression (unsurprisingly), but other than that I do not really know.

I also tried to see if equally interesting identities would emerge from the ordinary and exponential "generating function", but these end up having the form:

$$ \sum_{n=1}^{\infty} \frac{\ln(n)}{a^n} = \sum_{p} \ln(p) \left( \sum_{n=1}^{\infty} \frac{1}{a^{p^n}-1} \right) $$

$$ \sum_{n=1}^{\infty} \frac{\ln(n) a^n}{n!} = \sum_{p} \ln(p) \left( \sum_{n=1}^{\infty} \frac{1}{p^n} \cdot \left( \sum_{j=0}^{p^n-1} \left[\exp\left(e^{\frac{2\pi i}{p^n}}a\right)-1 \right] \right) \right) $$

So not as pretty nor as useful as we just get even more sums (I know technically $\zeta(s)$ is also a sum in disguise, but it is easier to evaluate than these other ones).

Anyways, since noone bothered to mention it anywhere does that mean this result is pretty unremarkable? Or have any of you seen it mentioned somewhere in some obscure book? Not that it matters since I think it is pretty cool, but normally I am able to find even the most obscure identities with enough searching.

Any other potential cool facts regarding any of the identities would also (of course) be appreciated. :)

  • $\begingroup$ I am sure that you know that the lhs is $-\zeta '(s)$ $\endgroup$ Jul 22, 2021 at 3:01

1 Answer 1


$\zeta(s)=\sum n^{-s}$, $\zeta'(s)=-\sum{\log n\over n^{-s}}$.

$\zeta(s)=\prod(1-p^{-s})^{-1}$, $\log\zeta(s)=-\sum\log(1-p^{-s})$, $${\zeta'(s)\over\zeta(s)}=-\sum{p^{-s}\log p\over1-p^{-s}}=-\sum{\log p\over p^s-1}$$

It's easy from there. The place to look for the formula is an Analytic Number Theory text that covers the zeta function.


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