how to estimate $\prod_{k=2}^n \log(k)$? I wonder if I can estimate $\prod_{k=2}^n \log(k)$ as $a^l$ for some a. I know that it is bounded by $e^{n^2}$, but I would like to get something finer. 
 A: $f(x)=\log\log x$ is a concave function on $[2,+\infty)$, since its derivative is $f'(x)=\frac{1}{x\log x}$ and its second derivative is $f''(x)=-\frac{1}{x^2}\left(\frac{1}{\log x}+\frac{1}{\log^2 x}\right)$. The Hermite-Hadamard inequality hence gives:
$$ \sum_{n=2}^{N}\log\log n \leq \int_{3/2}^{N+1/2}\log\log x\,dx = \left.\left(x\log\log x-\frac{x}{\log x}\right)\right|_{1/2}^{N+1/2}-\int_{1/2}^{N+1/2}\frac{dt}{\log^2 t}$$
and
$$ \exp\left(n\log\log n-\frac{n}{\log n}\right)$$
is a quite good approximation for $\prod_{k=2}^{n}\log k.$
A: Using geometric and arithmetic mean and Stirlings formula you get
$$
\left(\prod_{k=2}^n\ln(k)\right)^{1/(n-1)}\le\frac{\sum_{k=2}^n\ln(k)}{n-1}
=\frac{\ln(n!)}{n-1}\le\frac{\ln(\sqrt{2\pi(n+1)})+n·(\ln(n)-1)}{n-1}
$$
A: Using Abel's summation we have $$\sum_{n=2}^{N}\log\left(\log\left(n\right)\right)=\sum_{n=2}^{N}1\cdot\log\left(\log\left(n\right)\right)=\left(N-1\right)\log\left(\log\left(N\right)\right)-\int_{2}^{N}\frac{\left\lfloor t\right\rfloor -1}{t\log\left(t\right)}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function and using $\left\lfloor t\right\rfloor =t+O\left(1\right)
 $ we have $$=\left(N-1\right)\log\left(\log\left(N\right)\right)-\textrm{Li}\left(N\right)+O\left(\log\left(\log\left(N\right)\right)\right)
 $$ where $\textrm{Li}\left(N\right)$ is the logarithmic integral.
