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Let $1,2,3,4,5,6,7,8,9,11,12,\cdots$ be the sequence of all the positive integers which do not contain the digit zero. Write $\{a_n\}$ for this sequence. By comparing with a geometric series, show that

$$\sum_n \frac{1}{a_n}\lt90$$

I would try to start this but summing the geometric series is trivial, so the point lies in finding such a series(or a set of series)

This looks really amazing to me since this is just the harmonic series, with (apparently) a small fraction of the terms removed, and yet it converges.

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  • $\begingroup$ What is the question? $\endgroup$
    – user122283
    Apr 9, 2014 at 16:23
  • $\begingroup$ The number of terms in the sequence $a_n$ must be lesser than approximately $1.2204\times 10^{39}.$ $\endgroup$
    – user122283
    Apr 9, 2014 at 16:27
  • $\begingroup$ Here is an observation that I strongly suspect will lead to a solution (but I haven't got the time to actually try it): There are nine times as many $n+1$ digit numbers with no zeroes as there are $n$ digt numbers with no zeroes. You get the former by taking each of the latter, call it $k$, and forming $10k+1$, … ,$10k+9$. Now use the fact that each of the latter is $>10k$. I smell the sum $\sum(9/10)^n$ in there … $\endgroup$
    – Harald Hanche-Olsen
    Apr 9, 2014 at 16:27
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    $\begingroup$ This is a Kempner series. Duplicate of this, this, and this. $\endgroup$
    – Lucian
    Apr 9, 2014 at 16:32
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    $\begingroup$ Oh, and one more thing: You are not really removing a small fraction of the natural numbers. You are keeping $(9/10)^n$ of all $n$-digit numbers, which is a vanishingly small fraction when $n$ is large. $\endgroup$
    – Harald Hanche-Olsen
    Apr 9, 2014 at 16:32

1 Answer 1

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Let $S_1 = \sum\frac{1}{b_n}$ where $b_n$ is the finite subsequence of $a_n$ which are one digits numbers. Obviously $S_1$ is finite.

Let $S_2 = \sum\frac{1}{c_n}$ where $c_n$ is the finite subsequence of $a_n$ which are two digits numbers. Then $S_2 < \frac{9}{10}S_1$

Let $S_3 = \sum\frac{1}{d_n}$ where $d_n$ is the finite subsequence of $a_n$ which are three digits numbers Then $S_3$ is smaller than $\frac{9}{10}S_2$. Because of the fact that $\sum_{c=1}^9\frac{1}{abc} \leq \sum_{c=1}^{9}\frac{1}{ab0} \leq \sum_{c=1}^9\frac{1}{10}\frac{1}{ab}$.

Then the result follows if we compute $S_1$ as $\sum_{k=1}^9\frac{1}{k}$

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