$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots$$

Okay so we all know the harmonic series is divergent right? But apparently when you remove all the terms that has a nine in it, suddenly it becomes a convergent series.

Now this can easily be deduced via a geometric sum seen in https://en.wikipedia.org/wiki/Kempner_series .

$$\sum_{\text{$n$ does not contain $9$}} \frac{1}{n} < 8 \sum_{n = 1}^{\infty} \left(\frac{9}{10}\right)^{n-1} = 80.$$

What I find counter intuitive is the fact that when you literally take only 100 of the terms of the harmonic series (not arbitrarily of course) $$$$\frac 11\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots$$ \frac 1{100}\sum_{n=1}^\infty\frac1n=\frac 1 {100} + \frac 1{200} + \frac 1{300} + \cdots$$

i.e. 1/100 + 1/200 + 1/300 +..... or any number bigger than that, you get a series which is more than or equals to 1/100 multiplied by the harmonic series, which is infinity! So my real question is that is there a special way which the numbers are added or removed which affects the sum of the series? How do you identify/characterize this special form of choosing which numbers to include/exclude to form a convergent/divergent series?

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    $\begingroup$ On interesting note to your question about a "special way in which numbers are removed..." is the statement in your link about Fahri, who looked at the general case... $\endgroup$ – imranfat Oct 30 '15 at 15:55
  • $\begingroup$ Oh interesting, I did not realise farhi generalized it, my question was more on the phenomenon of removing 99/100 of the terms of the harmonic series (a.k.a. $\frac 1{100}\sum_{n=1}^\infty\frac1n=\frac 1 {100} + \frac 1{200} + \frac 1{300} + \cdots$) is still infinity while just removing all the terms with digit 9 yields a convergent series. So my question was, does the order of removal matter? $\endgroup$ – Zachary Nov 1 '15 at 3:47
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    $\begingroup$ I don't think so. The order of removal usually matters in case of an conditionally convergent series (think alternating) where different orders create different sums. In this case, all added terms are still positive and so if assuming that all terms in question are removed, I doubt that the sum will change if you don' t follow a particular order... $\endgroup$ – imranfat Nov 1 '15 at 16:38
  • $\begingroup$ I see, thank you for your insight in this matter $\endgroup$ – Zachary Nov 2 '15 at 10:54

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