How can i prove this inequality $\prod_{i=2}^n \log{i} \geq 3^{n+1} $?

Question

How can I prove this?
(This should be shown starting from $n_0 \geq 0$ s.t it's will be true for any $n \geq n_0 $, and need to find $K>0$ [while K is constant])

$$K*\prod_{i=2}^n \log{i} \geq 3^{n+1} $$

I don't have any idea, how to simplify this product..

the base of $log$ is 2.

Answer 1

For $n\geq 2^{2^3}=512$, we have

$$ \begin{array}{lcl} \sum_{k=2}^{n} \log(\log(k)) &=& \sum_{k=2}^{511} \log(\log(k))+\sum_{k=512}^{n} \log(\log(k)) \\ &\geq& \sum_{k=2}^{511} \log(\log(2)) +\sum_{k=512}^{n} \log(\log(512)) \\ &=&\sum_{k=2}^{511} 0+\sum_{k=512}^{n} 2\log(3) \\ &=& (n-511)(2\log(3)) \\ &=& (n+1+n-1023)\log(3) \\ \end{array} $$

We deduce that for $n\geq 1023$, we have $\sum_{k=2}^{n} \log(\log(k)) \geq (n+1)\log(3)$. QED

Answer 2

Straightforward induction: verify the inequality for $n$ for the smallest valid base case and then complete the induction for $n\geq7$. From the induction hypothesis, multiply both sides with $log_2 (n+1)$ and then use the fact that $log_2 (n+1) \geq 3$