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Math Help - Induction proof concerning a finite sum inequality

  1. #1
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    Post Induction proof concerning a finite sum inequality

    Here is the link:

    https://docs.google.com/viewer?a=v&p...thkey=CO2G4q4F

    Could someone help me with the style? Specifically, I don't like (4) and I wish there was a way to express (1) through a series of transformation on (3). Keeping the number of steps about the same or less of course.
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  2. #2
    Senior Member BAdhi's Avatar
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    here was the same post below. Sorry for the inconvenience
    Last edited by BAdhi; June 10th 2011 at 08:10 PM. Reason: deleted due to re-posting
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  3. #3
    Senior Member BAdhi's Avatar
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    this is how I got it,

    assume (1) is true up to n=p

    then,
    \sum_{i=0}^{p} \frac{1}{i^2} \leq 2-\frac{1}{p}

    adjust and subtract both sides by \frac{1}{(p+1)}
    \sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)}\leq 2-\frac{1}{(p+1)}

    since (8)
    \sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)}\geq \frac{1}{(p+1)^2}+\sum_{i=0}^{p} \frac{1}{i^2}
    \sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)}\geq \sum_{i=0}^{p+1} \frac{1}{i^2}
    \sum_{i=0}^{p+1} \frac{1}{i^2} \leq \sum_{i=0}^{p} \frac{1}{i^2}+\frac{1}{p}-\frac{1}{(p+1)} \leq 2-\frac{1}{(p+1)}

    \sum_{i=0}^{p+1} \frac{1}{i^2} \leq 2-\frac{1}{(p+1)} which prove correct for n=p+1
    Last edited by BAdhi; June 10th 2011 at 08:07 PM. Reason: latex error
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  4. #4
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    Try to mimick the style of typical proofs by induction:



    And in general, by mimicking the style of others one can eventually develop one's own style, or so I believe.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    A better bound:

     \sum_{i=0}^{p}\frac{1}{i^2}=1+\frac{1}{2^2}+\frac{  1}{3^2}+\frac{1}{4^2}+...+\frac{1}{p^2}<1+\frac{1}  {2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^p}=  2-(\frac{1}{2})^p
    Last edited by Also sprach Zarathustra; June 11th 2011 at 12:04 PM.
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  6. #6
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    Quote Originally Posted by Ontolog View Post
    Here is the link:

    https://docs.google.com/viewer?a=v&p...thkey=CO2G4q4F



    Could someone help me with the style? Specifically, I don't like (4) and I wish there was a way to express (1) through a series of transformation on (3). Keeping the number of steps about the same or less of course.
    Alternatively,

    P(k)

    \sum_{i=1}^{k}\frac{1}{i^2}\le\ 2-\frac{1}{k}


    P(k+1)

    \sum_{i=1}^{k+1}\frac{1}{i^2}\le\ 2-\frac{1}{k+1}

    Try to show that IF P(k) is true, THEN P(k+1) will also be true.

    If P(k) is true, then P(k+1) is

    x+\frac{1}{(k+1)^2}\le\ 2-\frac{1}{k+1}\;\;?

    where x\le\ 2-\frac{1}{k}

    x\le\ 2-\frac{1}{k+1}-\frac{1}{(k+1)^2}\;\;?\;\;\Rightarrow\ x\le\ 2-\left[\frac{k+1+1}{(k+1)^2}\right]\;\;?

    x\le\ 2-\left[\frac{1}{\left(\frac{[k+1]^2}{k+2}\right)}\right]\;\;?\;\;\Rightarrow\ x\le\ 2-\frac{1}{\left(\frac{k^2+2k+1}{k+2}\right)}\;\;?

    \Rightarrow\ x\le\ 2-\frac{1}{\left(\frac{k[k+2]+1}{k+2}\right)}\;\;?

    True, since

    \frac{k(k+2)+1}{k+2}>\frac{k(k+2)}{k+2}

    so the above line is subtracting a smaller value from 2 than 1/k
    Last edited by Archie Meade; June 11th 2011 at 09:14 AM.
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  7. #7
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    The inequality \sum_{i=1}^{n}\frac{1}{i^2}<2-\frac{1}{n} can be established without using induction.

    From i^2>(i-1)i, it follows \frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}<1  +\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{  1}{(n-1)\cdot n}, which is a telescoping sum.(a)
    Therefore, \sum_{i=1}^{n}\frac{1}{i^2}<1+1-\frac{1}{n}=2-\frac{1}{n}. (To Also sprach Zarathustra, 2-\frac{1}{n}<2-\frac{1}{2^n} for n\geq1)


    (a) \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1  }{(n-1)\cdot n}=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots+(\frac{1}{n-1}-\frac{1}{n}); the terms, except the first and last, cancel.
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