What's this partial sum $ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$ equal?

Question

I want to get this partial sum of $$ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$$ which it is convergent and it is closed to one half , I have tried to use polylogarithm function which is defined as : $\textrm{Li}_{s}\left(z\right)=\sum_{k\geq1}\frac{z^{k}}{k^{s}} $ using partial derivative such that for all complex $s$ and for $\left|z\right|<1 $. So observe that $$\frac{\partial}{\partial s}\left(\textrm{Li}_{s}\left(\frac{1}{2}\right)\right)=-\sum_{k\geq1}\frac{\log\left(k\right)}{2^{k}k^{s}} $$ but I can't come up to $\log(k!)$ in nominator any closed form for that partial sum , any help ?

Answer 1

You can do the following.

I'll call your partial sum $S_n$.

First you can get rid of the term $k=0$ since $\log 1 =0$. $$S_n = \sum_{k=0}^{n-1}\frac{\log(k!)}{2^{k+1}} = \sum_{k=1}^{n-1}\frac{\log(k!)}{2^{k+1}}\,.$$

Then you can use $$\log(k!) = \sum_{j=1}^k \log(j)\,.$$

You get $$S_n = \frac{\log(n-1)}{2^n} + \left(\frac{1}{2^n}+\frac{1}{2^{n-1}}\right)\log(n-2)+\dots$$

This becomes $$S_n = \frac{1}{2^n}\sum_{k=1}^{n-1}(2^k-1)\log(n-k) = \sum_{m=1}^{n-1}\left(\frac{1}{2^m}-\frac{1}{2^n}\right)\log m\,,$$ where $m=n-k$.

In this way you have got rid of the factorials.

Clearly $$\sum_{m=1}^{n-1} \log m = \log ((n-1)!)\,.$$ Mathematica tells me $$\sum_{m=1}^{n-1}\frac{\log m}{2^m} = 2^{-n}\, \mathrm{LerchPhi}^{(0,1,0)}(1/2,1,n)-\mathrm{PolyLog}^{(1,0)}(0,1/2)\,.$$