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Math Help - Series proof

  1. #1
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    Series proof

    Let p and n be real. Now show that
    <br />
n^p < \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^p \\<br />
    using the identity:
    <br />
b^n - a^n = (b-a)(b^{n-1} + b^{n-2}a + \dotsb + ba^{n-2} + a^{n-1}),  n \in \mathbb{N}<br />

    I have been trying at this one for hours, it is driving me nuts. Can someone point me in the right direction ?
    Last edited by hjortur; September 9th 2009 at 06:08 PM.
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  2. #2
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    Quote Originally Posted by hjortur View Post
    Let p and n be real. Now show that
    <br />
n^p \le \frac{(n+1)^{p+1} - n^{p+1}}{p+1} \le (n+1)^p \\<br />
    using the identity:
    <br />
b^n - a^n = (b-a)(b^{n-1} + b^{n-2}a + \dotsb + ba^{n-2} + a^{n-1}),  n \in \mathbb{N}<br />

    I have been trying at this one for hours, it is driving me nuts. Can someone point me in the right direction ?
    This only works for n,p\in\mathbb{N}.

    Notice that

    (n+1)^{p+1} - n^{p+1}=[(n+1)-n][(n+1)^p+(n+1)^{p-1}n+...+(n+1)n^{p-1}+n^p]

    =(n+1)^p+(n+1)^{p-1}n+...+(n+1)n^{p-1}+n^p.

    Now notice that each term in this series is represented by

    (n+1)^{p-i}n^i,

    and that there are a total of p+1 terms therein. Notice also that for each term, we have

    n^p\leq(n+1)^{p-i}n^i\leq(n+1)^p,

    which means that for the entire series we have

    (p+1)n^p\leq\sum_{i=0}^p(n+1)^{p-i}n^i\leq(p+1)(n+1)^p.

    Substituting, we have

    (p+1)n^p\leq(n+1)^{p+1} - n^{p+1}\leq(p+1)(n+1)^p.

    Then we divide by p+1:

    n^p\leq\frac{(n+1)^{p+1} - n^{p+1}}{p+1}\leq(n+1)^p. \blacksquare
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  3. #3
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    Yeah, sorry for that, I was mistranslating the word "real" .

    Also there was a typo (i am pretty new at this latex stuff) it was not supposed to be "less-than-or-equal" but "less-than". It is fixed above.

    Notice also that for each term, we have
    <br />
n^p\leq(n+1)^{p-i}n^i\leq(n+1)^p<br />
    so this one is supposed to be:
    <br />
n^p<(n+1)^{p-i}n^i<(n+1)^p<br />
    wich does not hold for i=0.

    Sorry for that.
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  4. #4
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    Quote Originally Posted by hjortur View Post
    Yeah, sorry for that, I was mistranslating the word "real" .

    Also there was a typo (i am pretty new at this latex stuff) it was not supposed to be "less-than-or-equal" but "less-than". It is fixed above.



    so this one is supposed to be:
    <br />
n^p<(n+1)^{p-i}n^i<(n+1)^p<br />
    wich does not hold for i=0.

    Sorry for that.
    The strict inequalities hold as long as i\neq 0 and i\neq p, which is satisfied for at least one term in the series so long as p\geq 2.
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