# Limit: ratio of the digit product and the number itself

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 $$n \ge 10^{k-1}$$
• @RossMillikan, is it: $$0\leq\lim_{n\to\infty}\frac{a_n}{n}\leq\lim_{k\to\infty}10\left(\frac{9}{10}\right)^k=0$$ – ms._VerkhovtsevaKatya Jan 28 at 8:43