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:
$$\boxed{
\;\;\;\;
\ln (n)=\sum_{k=1}^{\infty} \frac{\zeta(k+1)}{n^{k+1}} \sum_{m=1}^{n-1} m^{k}
\;\;\;\;}
$$
Best Answer
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}