Question: Can the following three conjectured closed form representations be proven to be correct assuming $\Re(s)>1$ and $m\in \mathbb{P}$?

$$\underset{N\to\infty}{\text{lim}}\left(-\sum\limits_{n=1}^N a(n) \frac{n\, m^{n\, s}}{\left(m^{n\, s}-1\right)^2}\right)=\frac{1}{1-m^s}\tag{1}$$

$$\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N \frac{a(n)}{m^{n s}-1}\right)=\log\left(\frac{1}{1-m^{-s}}\right)\tag{2}$$

$$\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{\phi(n)}{m^{n s}-1}\right)=\frac{1}{m^s+m^{-s}-2}\tag{3}$$

The arithmetic function $a(n)$ referenced on the left sides of formulas (1) and (2) above is defined as

$$a(n)=\frac{1}{n}\sum\limits_{d|n} \mu(d)\, d\tag{4}$$

and $\phi(n)$ referenced on the left side of formula (3) above is Euler's totient function.

Note that $a(n)\, n=\sum\limits_{d|n} \mu(d)\, d$ corresponds to OEIS entry A023900 Dirichlet inverse of Euler totient function (A000010) and

$$\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N A023900(n)\, n^{-s}\right)=\frac{\zeta(s)}{\zeta(s-1)},\quad\Re(s)\ge 2\tag{5}$$


$$\underset{\text{N}\to\infty}{\text{lim}}\left(\sum_{n=1}^N \phi(n)\, n^{-s}\right)=\frac{\zeta(s-1)}{\zeta(s)},\quad\Re(s)>2\tag{6}$$

I suspect the closed-form representations in formulas (1) to (3) above may actually be more generally valid for $\Re(s)>0$ and integer $m>1$ or perhaps even $m>1$, but my primary interest is the more limited case stated in the question above for reasons that will become apparent in the formulas below.

The left and right sides of formula (1) above are related to the following two formulas for $\frac{\zeta'(s)}{\zeta(s)}$

$$\frac{\zeta'(s)}{\zeta(s)}=\underset{M,N\to\infty}{\text{lim}}\left(-\sum\limits_{m=2}^M 1_{m\in\mathbb{P}}\, \log(m) \sum\limits_{n=1}^N a(n) \frac{n\, m^{n\, s}}{\left(m^{n\, s}-1\right)^2}\right),\quad\Re(s)>1\tag{7}$$

$$\frac{\zeta'(s)}{\zeta(s)}=\underset{M\to\infty}{\text{lim}}\left(\sum\limits_{m=2}^M 1_{m\in\mathbb{P}}\, \frac{\log(m)}{1-m^s}\right),\quad\Re(s)>1\tag{8}$$

where $1_{m\in\mathbb{P}}$ is the prime indicator function which returns the value $1$ if $m$ is a prime and zero otherwise.

The left and right sides of formula (2) above are related to the following two formulas for $\log\zeta(s)$:

$$\log\zeta(s)=\underset{M,N\to\infty}{\text{lim}}\left(\sum\limits_{m=2}^M 1_{m\in\mathbb{P}} \sum _{n=1}^N \frac{a(n)}{m^{n s}-1}\right),\quad\Re(s)>1\tag{9}$$

$$\log\zeta(s)=\underset{M\to\infty}{\text{lim}}\left(\sum\limits_{m=2}^M 1_{m\in\mathbb{P}}\, \log\left(\frac{1}{1-m^{-s}}\right)\right),\quad\Re(s)>1\tag{10}$$

The left and right sides of formula (3) above are related to the following two formulas for $K_\Omega(s)$

$$K_\Omega(s)=\underset{M,N\to\infty}{\text{lim}}\left(\sum\limits_{m=2}^M 1_{m\in\mathbb{P}}\, \sum\limits_{n=1}^N \frac{\phi(n)}{m^{n s}-1}\right),\quad\Re(s)>1\tag{11}$$

$$K_\Omega(s)=\underset{M\to\infty}{\text{lim}}\left(\sum\limits_{m=2}^M \frac{1_{m\in\mathbb{P}}}{m^s+m^{-s}-2}\right),\quad\Re(s)>1\tag{12}$$

which represent alternate ways to evaluate the Dirichlet series

$$K_\Omega(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N 1_{n=p^k}\, \Omega(n)\, n^{-s}\right),\quad\Re(s)>1\tag{13}$$

where $1_{n=p^k}$ is the prime-power indicator function which returns the value $1$ if $n$ is a prime-power and zero otherwise and $\Omega(n)$ counts the number of non-distinct primes in $n$ (i.e. $\Omega(n)=k$ when $n=p^k$).

I believe formula (13) for $K_\Omega(s)$ above can be analytically continued from $\Re(s)>1$ to $\Re(s)>0$ as follows

$$K_\Omega(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N b(n) \log(\zeta(n s))\right),\quad\Re(s)>0\tag{14}$$


$$b(n)=\frac{1}{n}\sum\limits_{d|n} d^2 \mu\left(\frac{n}{d}\right)\tag{15}$$

Note that $b(n)\, n=\sum\limits_{d|n} d^2 \mu\left(\frac{n}{d}\right)$ corresponds to OEIS Entry A007434 for the Jordan function $J_2(n)$ which is a generalization of Euler's totient function $\phi(n)$ and

$$\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \left(\sum\limits_{d|n} d^2 \mu\left(\frac{n}{d}\right)\right) n^{-s}\right)=\frac{\zeta(s-2)}{\zeta(s)},\quad\Re(s)>3\tag{16}$$

I believe formulas (7) to (12) above for $\frac{\zeta'(s)}{\zeta(s)}$, $\log\zeta(s)$, and $K_\Omega(s)$ are all valid for $\Re(s)>1$, but note formula (12) above for $K_\Omega(s)$ converges for $\Re(s)<-1$ as well as $\Re(s)>1$ and evaluates to $K_\Omega(-s)$=$K_\Omega(s)$ for $|\Re(s)|>1$.

  • $\begingroup$ Please define your notation $m\in \mathbb{P}$ at the beginning of your question. $\endgroup$
    – Somos
    Jun 16, 2022 at 20:53
  • $\begingroup$ @Somos $m\in \mathbb{P}$ is just a way of saying $m$ is an element of the prime numbers. My question was specific to $\Re(s)>1$ and $m\in\mathbb{P}$ because of the relationship of formulas (1) to (3) with formulas (7) to (12). I indicated in my question I suspect formulas (1) to (3) may be more generally valid for any $\Re(s)>0$ and perhaps $m>1$, but I now suspect they may be even more generally valid for $\Re(s)>0$ and $\Re(m)>1$. But this all based on a few evaluations of formulas (1) to (3), and I'm hoping someone can derive a proof of formulas (1) to (3). $\endgroup$ Jun 16, 2022 at 21:45
  • $\begingroup$ In your equations $(1),(2),(3)$ your could replace $m^s$ with $z\in\mathbb{C}$. $\endgroup$
    – Somos
    Jun 16, 2022 at 21:50
  • $\begingroup$ @Somos Perhaps that would be more general, but would it provide new insight perhaps leading to a proof? $\endgroup$ Jun 16, 2022 at 21:55
  • $\begingroup$ @Somos After replacing $m^s$ with $z$ in formulas (1) to (3), limited investigation seems to suggest the resulting formulas are all three valid for $|\Re(z)>1|$. $\endgroup$ Jun 16, 2022 at 22:39

1 Answer 1


For simplicity and generality make the definition $\,z := m^{-s}.\,$ Then your three equations become $$ -\sum_{n=1}^\infty a(n) \frac{n\,z^{-n}}{(z^{-n}-1)^2} =\frac{1}{1-z^{-1}},\tag{1}$$ $$ \sum_{n=1}^\infty \frac{a(n)}{z^{-n}-1} = \log\Big(\frac{1}{1-z}\Big), \tag{2}$$ $$ \sum_{n=1}^\infty \frac{\phi(n)}{z^{-n}-1} = \frac{z}{(1-z)^2}. \tag{3}$$ They all have radius of convergence $1$. For some background on series such as these refer to Wikipedia article Lambert series.

  • $\begingroup$ 3 clearly follows by expansion of LHS and grouping (allowed by absolute convergence on $|z|\le r<1$) as the generic term $z^m$ has coefficient $\sum_{d|m} \phi(d)=m$ which is the coefficient of RHS as that is the Koebe function; would expect 2 to follow the same way and then 1 follows from 2 by differentiation $\endgroup$
    – Conrad
    Jun 16, 2022 at 23:29

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