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Let ${\{n_1,n_2,…\} }$ be the set of natural numbers that do not use the digit 0 in their decimal expansion. Then, the series $\sum_{k=1}^\infty \frac{1}{n_k}$ converges to a number less than 90.

Is it ok to consider the series as $$\sum_{k=1}^\infty \frac{1}{n} - \sum_{i=1}^\infty \frac{1}{10^i}$$

This is an exercise problem in "Mathematical analysis" by ${Tom.M.Apostol.}$ I didn't understand the proof given for a similar question in the site already. I'm not good at permutations and combinations and inclusion and exclusion. Can anyone help with the proof with elementary calculations.

lulu
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Lakshmi Priya
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    See this https://math.stackexchange.com/a/364658/1093844 also you are omitting numbers like $10,100, \cdots$ but there are other numbers with $0$ in it like $20,105$ etc – Soumik Mukherjee Aug 09 '23 at 13:00
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    Note: your decomposition doesn't make sense. First of all, there are plenty of natural numbers like $10123$ that have a $0$ in their decimal expression but are not powers of $10$. Secondly, the first of sums diverges and you can't generally rearrange series if they diverge. – lulu Aug 09 '23 at 13:00
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    The main reason that approach won't work is that the harmonic series $\sum_{k=1}^\infty \frac{1}{n}$ diverges, so you would essentially get $\infty-\infty$ which is indeterminate. Note also that what you are subtracting is the reprocals of ${10^i}$, and this set excludes things like $20$, $501$, $23045$, etc. – Jaap Scherphuis Aug 09 '23 at 13:01
  • It's better to tell us which is the similar question on the site, and what it is about that solution that you don't understand. – Gerry Myerson Aug 09 '23 at 13:33

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