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Series and convergence

  1. Sep 4, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem states: In the harmonic series ##\sum_{1}^{\infty} \frac{1}{k}##, all terms for which the integer ##k## contains the digit 9 are deleted. Show that the resulting series is convergent.
    Hint: Show that the number of terms ##\frac{1}{k}## for which ##k## contains no nines and ##10^{p-1} \leq k < 10^p## is less than ##9^p##.
    2. Relevant equations


    3. The attempt at a solution
    Well, it is not hard to show and see the required in the hint, but other than that the sum of less terms will be smaller than the original number of terms in the harmonic series, I am not sure how I can deduce the convergence. Can't see any link to convergence.
    I would be happy to get help in this exercise.

    Thank you.
     
  2. jcsd
  3. Sep 4, 2016 #2

    PeroK

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    What are the simplest series you can think of?
     
  4. Sep 4, 2016 #3

    Mark44

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    I'm having a hard time understanding what this means. Is this the exact wording of the hint?
     
  5. Sep 4, 2016 #4
    well, I can think of ##\sum \frac{1}{k^2}##. Still can't see how to use it...

    Yes, those are the exact words. It means how many numbers there are between, for example 10 to 99(included), which do not have the digit 9. Example 19 is not counted, and also 95. So the hint says to show that there are, for example, between 10 to 99, less numbers without the digit 9 than 9^p, and not just between 10 to 99, but also between 100 to 999 and so on.
     
  6. Sep 4, 2016 #5

    Mark44

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    It was this part:
    I think what this is saying is that k < 9p, where ##10^{p-1} \leq k < 10^p##. As written, it suggests that ##10^p < 9^p##.
     
  7. Sep 4, 2016 #6

    PeroK

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    I assumed they were aiming for a geometric series.
     
  8. Sep 4, 2016 #7
    Thanks for the answers.

    I understood that the number of terms is less than 9^p, which is, well, true.

    Then I need a bit more help here.
     
  9. Sep 4, 2016 #8

    PeroK

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  10. Sep 4, 2016 #9

    Ray Vickson

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    For ##10^{p-1} \leq k < 10^p## you have no more than ##9^p## relevant terms, and each relevant term ##1/k## is ##\leq 10^{1-p}##, so what could you say about the sum of such ##1/k##?
     
    Last edited: Sep 4, 2016
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