# Representation of logarithms in terms of the zeta-$\zeta(n)$ function.

I've been looking for representations of logarithms in terms of the zeta function. So for $$\ln(2)$$ we have this and more here

\begin{align*} \ln(2)=&\sum_{k=1}^{\infty}\frac{\zeta(k+1)}{2^{k+1}}\\ \ln(2)=&1-\sum_{n=1}^{\infty} \frac{\zeta(2 n)}{2^{2 n-1}(2 n+1)}\\ \end{align*}

and for $$\ln(3)$$ I've found

\begin{align*} \ln(3)=1+2\sum_{k=1}^{\infty}\frac{\zeta(2k+1)}{3^{2k+1}} \end{align*}

but for other logarithms,$$\ln(4)$$, $$\ln(5)$$ etc.., I didn't find anything.

Could someone point directions or post here representations of logarithms in terms of $$\zeta(n)$$?

Thanks.

EDIT: After some refinement of the expression given bellow I was able to get this beauty: $$\boxed{ \;\;\;\; \ln (n)=\sum_{k=1}^{\infty} \frac{\zeta(k+1)}{n^{k+1}} \sum_{m=1}^{n-1} m^{k} \;\;\;\;}$$

We use the series (DLMF) $$\begin{equation} \sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}z^{k}=-\gamma z+\ln\Gamma\left (1-z\right) \end{equation}$$ valid for $$\left|z\right|<1$$ and the multiplication formula for the gamma function (DLMF) \begin{align} \prod_{k=1}^{n-1}\Gamma\left(\frac{k}{n}\right)=(2\pi)^{(n-1)/2}n^{-1/2} \end{align} Taking the logarithm and reversing the order of summation of the later expression gives $$\begin{equation} 2\sum_{p=1}^{n-1}\ln\Gamma\left(1-\frac{p}{n}\right)=(n-1)\ln(2\pi)-\ln n \end{equation}$$ From these results, it comes $$\begin{equation} \ln n=(n-1)\left(\ln(2\pi)-\gamma \right)-2\sum_{k=2}^\infty \frac{\zeta(k)}{kn^k}\sum_{p=1}^{n-1}p^k \end{equation}$$ If we want to remove the $$\ln(2\pi)-\gamma$$ term, we can use the expression for $$n=2$$: $$\begin{equation} \ln 2=\ln(2\pi)-\gamma -2\sum_{k=2}^\infty \frac{\zeta(k)}{k2^k} \end{equation}$$ and the classical expansion: $$\begin{equation} \ln(2)=\sum_{k=2}^{\infty}\frac{\zeta(k)}{2^{k}} \end{equation}$$ to deduce \begin{align} \ln(2\pi)-\gamma& =\sum_{k=2}^{\infty}\frac{\zeta(k)}{2^{k}}+2\sum_{k=2}^\infty \frac{\zeta(k)}{k2^k}\\ &=\sum_{k=2}^{\infty}\frac{\zeta(k)}{2^{k}}\left( 1+\frac{2}{k} \right) \end{align} and thus $$\begin{equation} \ln n=(n-1)\sum_{k=2}^{\infty}\frac{\zeta(k)}{2^{k}}\left( 1+\frac{2}{k} \right)-2\sum_{k=2}^\infty \frac{\zeta(k)}{kn^k}\sum_{p=1}^{n-1}p^k \end{equation}$$
• Alternatively from $\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}z^{k}=-\gamma z+\ln\Gamma\left(1-z\right)$ use $\log\Gamma(1+1/n) = -\log n +\log \Gamma(1/n)$ which gives $-\log n+\gamma$ taking $n=2$ and the series for $\log 2$ gives a series for $\gamma$ thus $\log n$ – reuns Dec 2 '19 at 3:59
Note that, $$ln(4) = 2ln(2)$$. Since you have the representation for $$ln(2)$$, you can obtain one for $$ln(4)$$. So in general, for $$ln(2^k)$$ as well, where $$k=1,2,3,4...$$.
Similarly you can find the representation for any integer $$n$$ of the form $$n={2^a}{3^b}$$, where $$a,b$$ are positive integers