TITLE: 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 [0 upvotes]: 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 $f(n)$

REPLY [1 votes]: 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 $3^9$ ?
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