3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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}

$\endgroup$
1
  • 2
    $\begingroup$ 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 $\log n$ $\endgroup$ – reuns Dec 2 '19 at 3:59
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.