# Formula for Sum of Logarithms $\ln(n)^m$

As you know $\sum_{n=1}^k \ln(n) =\ln(k!)$ is there a formula for $\sum_{n=1}^k \ln(n)^m$?

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Perhaps an idea - not sure if it will help. $$\sum \ln^2 n = \sum \ln n^{\ln n} = \ln \left( \prod n^{\ln n} \right).$$ –  gt6989b May 17 at 14:27
If m is constant, then $\sum_{n=1}^{k}ln(n)^m=\sum_{n=1}^{k}mln(n)=mln(k!)$ –  Noy Soffer May 17 at 14:30
look for (ln(n))^m –  Keyvan Ghaffari May 17 at 14:39
it ln(n)^m not ln(n^m) –  Keyvan Ghaffari May 17 at 14:47
m is a natural number –  Keyvan Ghaffari May 17 at 14:51
\begin{align} \sum_{n=1}^k\log(n)^m &\sim k\left(\log(k)^m-m\log(k)^{m-1}+m(m-1)\log(k)^{m-2}-\dots+(-1)^mm!\right)\\ &+\frac12\log(k)^m+C+\frac{m}{12k}\log(k)^{m-1}+O\left(\frac{\log(k)^{m-1}}{k^3}\right) \end{align} The constant $C$ depends on $m$ and needs to be determined separately. For $m=1$, Stirling's approsimation says that $C=\frac12\log(2\pi)$.
I would doubt there is an exact formula for $m\gt1$, but that is not to say there might not be one. –  robjohn May 23 at 1:53
@KeyvanGhaffari: Yes. For $m=1$, this is Stirling's approximation. The error term, $O\left(\dfrac{\log(k)^{m-1}}{k^3}\right)$, gets smaller as $k$ gets bigger. As I said, I doubt there is an exact formula for $m\gt1$. The exact formula for $m=1$ exists only because we've defined $n!=n(n-1)(n-2)\dots1$ –  robjohn May 23 at 12:17