Compute:$$\lim_{n\to\infty}\frac{a_n}{n},a,n\in\mathbb N$$ Where $a_n$ equals the product of the digits of $n$ in base $10$.
source Math Analysis 1 exam, 2012
My attempt: The first idea that came to my mind: $$a_n=0\;\forall\;n,k\in\mathbb N, s.t. n=10k$$ There certainly are some convergent subsequences $(a_{p(n)})$: $$\lim_{n\to\infty}a_{p(n)}=0$$
I thought of writing $n$ either in this polynomial form: $$n=\sum_{i=0}^nd_i10^i$$ or in Horner's algorithm: $$\begin{align*} n&=d_{0}+10\left(d_{1}+10\left(d_{2}+10\left(d_{3}+\cdots+10\left(d_{k-1}+10 d_{k}\right) \cdots\right)\right)\right),\\ k&=\lfloor\log n\rfloor+1 \end{align*}$$ (In my country we denote Briggs logarithm with $\log$)
The first option seemed better.
Then I decided to express $a_n$ this way: $$\prod_{i=0}^{\lfloor\log n\rfloor}d_i$$ I got this: $$\lim_{n\to\infty}\frac{\displaystyle\prod_{i=0}^{\lfloor\log n\rfloor}d_i}{\displaystyle\sum_{i=0}^nd_i10^i}$$ but I stuck here not knowing how to write a concise proof. Is there a better way of solving this?
Hint: for a number with $k$ digits, $a_n \le 9^k$ while $n \ge 10^{k-1}$