I have a summation series that unfortunately involves a log of a log. It looks like the following (assume all $\log$ are log base $2$):
$$ \sum_{i=1}^k \log\log\frac{n}{2^{k-i}} $$
I'd like to simplify this somehow, such as move the logs or the n value to the outside of the equation, so that I can use the closed form of the geometric series sum to further simplify the equation, but I'm not sure where to start.
Thanks for the help!
Well, let's see: $$\sum_{i=0}^{k-1} \log(\log(n\cdot 2^{-i})) = \log\left(\prod_{i=0}^{k-1}\log n + \log 2^{-i}\right) = \log\left(\prod_{i=0}^{k-1} \log n - i\right) = \log\Big(k! \binom{\log n}k\Big) = \log((\log n)_k) = \log\Big(\frac{\Gamma(\log n +1)}{\Gamma(\log n - k + 1)}\Big)=\color{maroon}{\log(\Gamma(\log n + 1)) - \log(\Gamma(\log n - k + 1))}$$ Where $(x)_k$ is the Pochhammer symbol $$(x)_k = x(x-1)\cdots(x-k+1)$$ We used $\log a + \log b = \log(ab)$ and $\log_2 2^k = k$.