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Prove that the sum of reciprocals of integers with no zeros in their decimal representation converges

Consider the positive integer with no zeros in their decimal representation:

    \[ n_1 = 1, \ n_2 = 2, \ \ldots \ n_9 = 9, \ n_{10} = 11, \ldots, n_{18} = 19, \ n_{19} = 21, \ldots. \]

Prove that the series

    \[ \sum_{n=1}^{\infty} \frac{1}{n_k} \]

converges. Further, prove that the sum is less than 90.


Incomplete.

2 comments

  1. Rafael Deiga says:

    Solution:
    First, note that 1/b<(1/10)^(k-1) for all b with k digits (and without the digit 0). Furthermore, there are exactly 9^k numbers b with k digits, because there are nine possibilites for each digit. Therefore, the sum of all b's is smaller than 9*(9/10)^(k-1). Thus, 1+1/2+1/3+1/4+… < 9*(1+9/10+(9/10)^2+…), which converges to 90.

    Sorry about my solution, I don't know how to use latex here.

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