Consider the positive integer with no zeros in their decimal representation:
![\[ n_1 = 1, \ n_2 = 2, \ \ldots \ n_9 = 9, \ n_{10} = 11, \ldots, n_{18} = 19, \ n_{19} = 21, \ldots. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-627215813ed4609c913de6a3c389116b_l3.png "Rendered by QuickLaTeX.com")
Prove that the series
![\[ \sum_{n=1}^{\infty} \frac{1}{n_k} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3efa43154ded655e8637caf80b927814_l3.png "Rendered by QuickLaTeX.com")
converges. Further, prove that the sum is less than 90.
Incomplete.
Consider the positive integer with no zeros in their decimal representation:
Prove that the series
converges. Further, prove that the sum is less than 90.
Incomplete.
These seem to be Kempner series: https://en.m.wikipedia.org/wiki/Kempner_series
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.
Thank you!!!
This is very good. But there is a small mistake, 1/b<(1/10)^(k-1) is not true for b = 1, k=1. But if you change it to <= instead of <. Unfortunately this does not guarantee that the sum is less than 90 only that the sum is less than or equal to 90. To prove that it is less than 90 write out the first ten terms and sum them up and add the rest with the geometric series formula.