Find minimum of $f(n)$ where $f(n)=\frac{\log_{3}{2}\log_{3}{3}\log_{3}{4}\cdots \log_{3}{n}}{9^n}$

Question

Find minimum of $f(n)$ where $$f(n)=\frac{\log_{3}{2}\log_{3}{3}\log_{3}{4}\cdots \log_{3}{n}}{9^n}, \qquad n\in \mathbb N$$

I not know how to transform the expression so that I find minimum of $f(n)$

Answer 1

Note $$f(n+1)=\frac{f(n)\log_3(n+1)}{9} \ \ \ \ \ \ \ \ \ \ (\text{i}) $$ $f(n)$ is decreasing for small $n$.

If $f(n)$ is a minimum, $f(n+1) \geq f(n) \implies \log_3(n+1) \geq 9 \implies n \geq 3^9-1$.

We see that $$ f(3^9-1) = f(3^9) $$Using (i) Can you prove that $f$ is strictly decreasing till $3^9-1$ and strictly increasing from $3^9$ ?