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Math Help - stuck on proving properties of a series

  1. #1
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    stuck on proving properties of a series

    i am stuck on proving this property of a series:

    let p_n=sigma (from k=1 to n) (1/k) and p=sigma (from k=1 to infinite) (1/k).

    Prove for all N, a natural number, there exists a n_0 in the naturals such that abs (p_n-p_n-1)<=1/N for all n>=n_0 yet abs (p-p_n)> 1/N for all n>=n_0.

    essentially one would need to show that abs (1/n) <= 1/N and that sigma(from k=n+1 to infinite) (1/n)>=1/N. how would we choose our n_0 in a way that this would work? thanks!
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  2. #2
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    never mind, just sparked in my head that this is a very simple induction problem for cases where n_0 is greater than or equal to 2. =)
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  3. #3
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    i was trying to show that this inequality holds for n+1 case, but found out that theres no way to bound this by the n case which we can assume true by induction hypothesis. Help? thanks. ((
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    Quote Originally Posted by squarerootof2 View Post
    i am stuck on proving this property of a series:

    let p_n=sigma (from k=1 to n) (1/k) and p=sigma (from k=1 to infinite) (1/k).

    Prove for all N, a natural number, there exists a n_0 in the naturals such that abs (p_n-p_n-1)<=1/N for all n>=n_0 yet abs (p-p_n)> 1/N for all n>=n_0.

    essentially one would need to show that abs (1/n) <= 1/N and that sigma(from k=n+1 to infinite) (1/n)>=1/N. how would we choose our n_0 in a way that this would work? thanks!
    Hmm well as far as I can see n_0 = N, works wonderfully well
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  5. #5
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    i can see that we are going to need proof by induction, but can you find a way to prove the second condition for the n+1 case? i cant seem to find a way that works. - thanks
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  6. #6
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    Quote Originally Posted by squarerootof2 View Post
    i am stuck on proving this property of a series:

    let p_n=sigma (from k=1 to n) (1/k) and p=sigma (from k=1 to infinite) (1/k).
    Am I missing something here?

    You define:

    p=\sum_{k=1}^{\infty} \frac{1}{k}

    That is p is the sum of the harmonic series, but the harmonic series is divergent!

    Ahh... now I see what is going on, you are proving that the harmonic series is divergent by contradiction... or meybe not

    RonL
    Last edited by CaptainBlack; July 5th 2008 at 02:27 AM. Reason: answer own question
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  7. #7
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    Quote Originally Posted by squarerootof2 View Post
    i am stuck on proving this property of a series:

    let p_n=sigma (from k=1 to n) (1/k) and p=sigma (from k=1 to infinite) (1/k).

    Prove for all N, a natural number, there exists a n_0 in the naturals such that abs (p_n-p_n-1)<=1/N for all n>=n_0 yet abs (p-p_n)> 1/N for all n>=n_0.

    essentially one would need to show that abs (1/n) <= 1/N and that sigma(from k=n+1 to infinite) (1/n)>=1/N. how would we choose our n_0 in a way that this would work? thanks!

    Let:

    p_n=\sum_{k=1}^n \frac{1}{k}

    and suppose:

    p_n=\sum_{k=1}^{\infty} \frac{1}{k}

    exists.

    Then for any integer n:

    p_n-p_{n+1}=\frac{1}{n}

    so for any n\ge N:

    p_n-p_{n+1}=\frac{1}{n} \le \frac{1}{N}.

    Now consider:

     <br />
p-p_n=\sum_{k=n+1}^{\infty}\frac{1}{k}=\sum_{k=n+1}^  {2N}\frac{1}{k}+\sum_{k=2n+1}^{4n}\frac{1}{k}+\sum  _{k=4n+1}^{\infty}\frac{1}{k}<br />

    The first series comprises n terms all greater than or equal to 1/(2n) so is greater than or equal to 1/2, and the second series is similarly greater than or equal to 1/2 and the third is greater thab zero, so:

     <br />
p-p_n=\sum_{k=n+1}^{\infty}\frac{1}{k} \ge 1 \ge \frac{1}{N}<br />

    So putting n_0=N will do.

    However note that this is all predicated on the existance of p, but there is no such p as the harmonic series diverges.

    RonL
    Last edited by CaptainBlack; July 5th 2008 at 05:40 AM.
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  8. #8
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    one question, do u mean (1/2)n instead of 1/(2n) when u justify the first series being greater than 1/2? and also, this is from the definition of harmonic series, right?
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  9. #9
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    Quote Originally Posted by squarerootof2 View Post
    one question, do u mean (1/2)n instead of 1/(2n) when u justify the first series being greater than 1/2? and also, this is from the definition of harmonic series, right?
    No I mean 1/(2n), since this is the samllest term in that finite series, and by construstion there are n terms all greater than or equal to the smallest.

    No I am using the series given, it just happens to be the harmonic series

    This is an adaptation of one method of proving that the harmonic series is divergent.

    RonL
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