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Reciprocal Sum

  1. May 22, 2005 #1
    I want to find the sum of the harmonic series where the n (as in SIGMA 1/n -- sorry for not using latex, the preview post button keeps displaying the wrong math symbols) cannot be a number that uses the digit 0.

    I've thought about doing a direct comparison test, comparing the sum to something like SIGMA (1/n - 1/(10^n)). I plan to subtract only the powers of ten or other numbers involving zero (e.g., 10*n) from the harmonic series, and see if that sum converges. But I'm faced with two problems:

    1. I can't seem to find the appropriate sum to subtract from the harmonic series
    2. I don't know how to find the sum to which a series (other than a Taylor series) converges.

    Can someone help? Thanks.

    P.S. SIGMA really means sigma from n = 0 to infinity
     
  2. jcsd
  3. May 22, 2005 #2

    Zurtex

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    The sum clearly does not converge if that helps, 1/10^n gets far too small too quickly, think about that then try and apply something mathematical :shy:. But also you should note that it is the sum from n=1 to infinity, as you have 1/n.
     
  4. May 22, 2005 #3
    Oops. Right, from n = 1 to infinity. Thanks for the help, though.

    So I need something that's big enough to slow 1/n down to something that converges... can someone help me find it?
     
  5. May 22, 2005 #4

    shmoe

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    To clarify, you're looking at the harmonic series where all terms with the digit 0 are removed:

    1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/9+1/11+1/12+...+1/19+1/21+...

    correct?

    Are you just looking to prove convergence or divergence or do you actually want to find the sum (if it does converge)?
     
  6. May 22, 2005 #5
    Correct. I am actually seeking to prove that the sum converges to a number less than 90.
     
  7. May 22, 2005 #6

    shmoe

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    Ok, actually finding what it converges to is a bit more work! A hint then on how to group the terms for proving your bound:

    (1/1+...+1/9)+(1/11+...+1/99)+(1/111+...1/999)+...
     
  8. May 22, 2005 #7
    Thanks. But that brings up two more questions:

    1. Where can I find (online) resources that tell me how to find what non-Taylor series converge to? (Unless, of course, the series is glaringly obvious)

    2. When grouping (1/1 + ... 1/9) + (1/11 + ... 1/99) + (1/111 + ... 1/999), how should I express it mathematically? (SIGMA 1/n - 1/(10^n) comes to mind, but as Zurtex said, 1/(10^n) is converges too quickly.) Should I use some sort of nested sums? (I.e., a SIGMA within a SIGMA)
     
  9. May 22, 2005 #8

    shmoe

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    In general it can be difficult to know what a series converges to. I've never seen the one you're working with here expressed in a neat form (like you can with geometric series or say [itex]\sum_{n=1}^{\infty}1/n^2[/itex]), only a computation of the first x number of digits. For a pretty classical problem like this one, entering the right things in google will turn up results.

    "SIGMA 1/n - 1/(10^n)" this seems worthless here. For a start, *most* numbers with a zero digit are not of the form 10^n.

    You can use a nested sum if you can manage a clean enough notation, but writing out the first few groupings and explaining in words what you're doing might be sufficient. Depends on who's reading it in the end.
     
  10. May 23, 2005 #9

    Zurtex

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    It's simpler than that, first of all, all terms are positive! This is quite important because if it converges, it absolutely converges, which allows us to rearrange the terms:

    [1/2] + [(1/3 + 1/4) - 1/10] + [(1/5 + 1/6 + 1/7 + 1/8) - 1/100] + ...

    Notice all the terms in the square brackets are greater or equal to 1/2. To prove that it does not converge to something less than 90, calculate the following:

    [tex]\sum_{n=0}^{2^{1000}} \left( \frac{1}{n} - \frac{1}{10^n} \right) \approx 760.825[/tex]

    This is not a proof, you will need to work on it to prove that everything I've said above is true.

    As for non-taylor series, or even taylor series, you might want to look at a program called Mathematica, it can perform many mathematical operations such as that symbolically or at least approximate them.
     
  11. May 23, 2005 #10

    shmoe

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    [itex]\sum\left(\frac{1}{n}-\frac{1}{10^n}\right)[/itex] is not the sum he's really interested in, it was just one he wanted to compare his series with (though that won't work).

    Even under the assumption that [itex]\sum\left(\frac{1}{n}-\frac{1}{10^n}\right)[/itex] is convergent (false of course), what you've done is not justified. You didn't just rearrange the terms, you've broken them apart.
     
  12. May 23, 2005 #11

    Zurtex

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    Are you sure? Oh well, shows how much I know about series. I thought it was valid as if it was convergent it would not be a conditionally convergent series and so I could do what I did as it didn't fall under the problem of the Riemann Series Theorem and I never lost any of the terms in the whole sum.
     
  13. May 23, 2005 #12

    shmoe

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    You are right that if it's convergent then it's absolutely convergent (terms all positive) so rearragning the terms would be fine.

    You didn't rearrange the terms though. The series is [itex]\sum a_n[/itex] where [itex]a_n=\frac{1}{n}-\frac{1}{10^n}[/itex]. A rearrangement needs a bijection f from the naturals to the naturals to get a sum like [itex]\sum a_{f(n)}[/itex]. Try finding an f that gives your new series.

    You thought of it as a sum like [itex]\sum (b_n-c_n)[/itex] and rearranged the b's and c's as you saw fit, which is not always valid. Consider the convergent sum [itex]\sum\left(\frac{1}{n}-\frac{1}{n+1}\right)[/itex]. You should be able to arrange the terms 1/1, 1/2, 1/3 ... and -1/2, -1/3, -1/4, ... into any sum you like (divergent or otherwise).
     
    Last edited: May 23, 2005
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