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 $n \ge 10^{k-1}$

  • 1
    $\begingroup$ Is the squeeze theorem suitable in this case? $\endgroup$ – Orchid_2.718281828 Jan 27 at 20:29
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    $\begingroup$ @Praskovya2.718281828 Yes, it is $\endgroup$ – Ross Millikan Jan 27 at 23:42
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    $\begingroup$ @RossMillikan, is it: $$0\leq\lim_{n\to\infty}\frac{a_n}{n}\leq\lim_{k\to\infty}10\left(\frac{9}{10}\right)^k=0$$ $\endgroup$ – ms._VerkhovtsevaKatya Jan 28 at 8:43
  • $\begingroup$ @VerkhovtsevaKatya: That is correct. $\endgroup$ – Ross Millikan Jan 28 at 14:01

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