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It's easy to see that for $Re(s)>0$,

$$\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{k=0}^{\infty}(-1)^k\frac{\ln^k(n)s^k}{k!}$$

from the series definition of the Riemann Zeta function and the Maclaurin series for $n^{-s}$ in terms of $s$. This gives the series representation of the coefficient for the $k$th term (i.e., the constant by which the term $s^k$ is scaled):

$$(-1)^k\frac{1}{k!}\displaystyle\sum_{n=1}^{\infty}\ln^k(n)$$

This begs the immediate question,

$$\displaystyle\sum_{n=1}^{\infty}\ln^k(n)=?$$

It should be noted that this series is divergent, but any way to analytically find an answer that would make sense would be great, like something along the lines of the method by which $\zeta(0)$ or the p-adic definition of $\sum_{n=0}^{\infty}2^n$ is found.

If you are not able to answer this question but are able to give me a representation of the Maclaurin series of the Riemann Zeta function without a double summation, that would be very nice too.

46andpi
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  • the identity, as stated, is not true. It must be said that it holds only for $\Re(s)>1$, otherwise you can't use the Dirichlet series to define the Riemann $\zeta$ function. – Masacroso Jun 18 '18 at 20:35
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    The sum diverges. The terms do not converge to zero. – Ross Millikan Jun 18 '18 at 20:56
  • The Riemann-Zeta function has a pole at $s=1$ – Mark Viola Jun 18 '18 at 21:40
  • Thanks for the comments, guys. I've made edits to make my question more intelligible. One minor question I have is how to edit the series to adapt for the pole at $s=1$, if any edits are necessary. As for the question being marked as a duplicate, I feel that it's important to note that I'm specifically requesting a form that can be expressed with a single summation. If the user that marked it as a duplicate still believes that the question is not sufficiently unique, I'll personally take it down. – 46andpi Jun 19 '18 at 01:35

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