Finding lower/upper bounds for $\prod_{i=2}^n \log(i)$ I have a homework problem where I need to asymptotically order a set of functions, and $\prod_{i=2}^n \log(i)$ is one of them.
Is there a tight upper/lower bound for this function?
I've tried the obvious upper bound of $\prod_{i=2}^n \log(n)$, and I've tried $2^n$ as a lower bound, but it's not particularly tight..
 A: One quick lower bound is
$$ \prod_{i=2}^n \log(i) \ge \prod_{i=n/2}^n \log(i)
\ge (\log(\tfrac n2))^{n/2} $$
which is like $c^{n\log\log n}$, matching your upper bound at this coarse level of comparison, and better than $c^n$.
A: By partial summation $$\sum_{i=2}^{n}\log\left(\log\left(i\right)\right)=n\log\left(\log\left(n\right)\right)-\int_{2}^{n}\frac{\left\lfloor t\right\rfloor }{t\log\left(t\right)}dt
 $$ where $\left\lfloor t\right\rfloor =t-\left\{ t\right\} 
 $ is the floor function and $\left\{ t\right\} 
 $ is the sawtooth function. Using the bounds $0\leq\left\{ t\right\} <1
 $ we get $$n\log\left(\log\left(n\right)\right)-\textrm{Li}\left(n\right)\leq\sum_{i=2}^{n}\log\left(\log\left(i\right)\right)\leq (n+1)\log\left(\log\left(n\right)\right)-\textrm{Li}\left(n\right)-\log\left(\log\left(2\right)\right)
 $$ where $\textrm{Li}\left(n\right)=\int_{2}^{n}\frac{1}{\log\left(t\right)}dt
 $ is the logarithmic integral. Finally $$\frac{\log^{n}\left(n\right)}{e^{\textrm{Li}\left(n\right)}}\leq\prod_{i=2}^{n}\log\left(i\right)\leq\frac{\log^{n+1}\left(n\right)}{e^{\textrm{Li}\left(n\right)}\log\left(2\right)}.
 $$
A: This is not an answer but it is too long for a comment (hoping it will give you some ideas).
Elaqqad gave a very interesting suggestion $$\log(P_n)\sim\int_{e}^n\log(\log(x))dx=\text{Ei}(1)-\text{li}(n)+n \log (\log (n))$$ where appear both the logarithmic and the exponential integrals. For $n=1000$, the approximation gives $1756.93$ for an exact value of $1757.54$; for $n=10000$, the approximation gives $20959.0$ for an exact value of $20959.8$.
On the other hand, if $\log(P_n)$ is plotted as a function of $n$, it seems that a rather good approximation (least square fit) is given by $$\log(P_n)\approx 1.2156 \,n^{1.05957}-70.4864$$ which gives values of $1763.95$ for $n=1000$ and $20970.8$ for $n=10000$.
A: With the help of the Abel–Plana formula (cf. http://dlmf.nist.gov/2.10.E2), it can be shown that
\begin{align*}
\prod\limits_{k = 2}^n {\log k}  =\; & C(\log n)^{n + 1/2} \exp \!\left( { - \int_2^n {\frac{{\text{d}t}}{{\log t}}} } \right) \\ & \times\!\left( 1 + \frac{1}{{12n\log n}} + \frac{1}{{288n^2 \log ^2 n}} +  \mathcal{O}\!\left( {\frac{1}{{n^3 \log n}}} \right) \right)
\end{align*}
with $C = 1.6352358009117223 \ldots$, or explicitly
$$
\log C =  - \frac{3}{2}\log \log 2 - 2\int_0^{ + \infty } {\frac{1}{{\text{e}^{2\pi t}  - 1}}\arctan \left( {\frac{{2\arctan (t/2)}}{{\log (t^2  + 4)}}} \right)\text{d}t} .
$$
