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The exercise $10.22.16$ goes as follows:

Let $n_1 < n_2 < n_3 ...$ denote positive integers that do not involve digit 0 in their decimal representation. Thus $n_1 = 1, n_2 = 2, ..., n_9 = 9, n_{10} = 11, ..., n_{18} = 19, n_{19} = 21,$ etc. Show that the series of reciprocals $\sum_{k=1}^{\infty} \frac{1}{n_k}$ converges.

When proving Riemann's rearrangement theorem, we were taking in order positive elements, and then in order negative elements, and iterating like that, and that was fine, because I thought we can do that because we were taking positive and negative elements in order and the sequence is conditionally convergent. However, if I apply a similar approach here (which maybe we could see like iteratively taking in order $10$ elements of $\frac{1}{k}$ sequence, and then the next negative element from the $\frac{1}{10k}$ sequence), I get:

$S = \sum_{k=1}^{\infty} \frac{1}{n_k} = \sum \frac{1}{k} - \sum \frac{1}{10 k} = \sum (\frac{1}{k} - \frac{1}{10 k}) =\frac{9}{10} \sum \frac{1}{k}$

Because we know that $\sum \frac{1}{n}$ diverges, from that I conclude that the series $S$ also diverges, which is a contradiction.

Can you please point out a mistake in my reasoning and a justification why that's incorrect?

I suspect it's because the $2$ harmonic series I was summing and rearranging are not convergent (while in the Riemann theorem's proof the sequence being rearranged is convergent), but I don't know why I could not combine them like I did above.

S11n
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    $\infty-\infty$ is an indeterminate form. For a correct solution, see Sum of reciprocals of numbers with certain terms omitted – Anne Bauval Nov 14 '22 at 11:15
  • @AnneBauval Thanks! Could you please explain in a full answer where that indeterminate form jumps in and why it's indeterminate? At this point I'm not looking for a solution to the exercise itself, but rather to understand where I derailed when I arrived at the contradiction... – S11n Nov 14 '22 at 11:53
  • @AnneBauval Actually, could it be that the problem is that $\sum_{k=1}^{\infty} \frac{1}{n_k} \ne \sum \frac{1}{k} - \sum \frac{1}{10 k}$, because the removed terms are not only 10k, but also for example $101, 102, ..., 1001, 1002, ...$? If we would just remove the $10, 20$, then it seems that such a sequence would still diverge, as I've shown, but that's a different sequence from the one I was supposed to solve... – S11n Nov 14 '22 at 12:10
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    The LHS is $\ne$ from the RHS above all because the RHS is $\infty-\infty.$ – Anne Bauval Nov 14 '22 at 12:20
  • @AnneBauval Thanks! Can you please make that a full answer, so I can close the question as answered? – S11n Nov 14 '22 at 14:22

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The problem in my reasoning above was that $∑1/_$ ≠ ∑1/− ∑1/10, because the removed terms are not only 10k, but also for example 101,102,...,1001,1002,...

For a correct answer to the Kempner sum, please see comments.

S11n
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