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Thread: Series proof

  1. #1
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    Series proof

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

    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; Sep 9th 2009 at 05:08 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
    $\displaystyle
    n^p \le \frac{(n+1)^{p+1} - n^{p+1}}{p+1} \le (n+1)^p \\
    $
    using the identity:
    $\displaystyle
    b^n - a^n = (b-a)(b^{n-1} + b^{n-2}a + \dotsb + ba^{n-2} + a^{n-1}), n \in \mathbb{N}
    $

    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 $\displaystyle n,p\in\mathbb{N}$.

    Notice that

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

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

    Now notice that each term in this series is represented by

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

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

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

    which means that for the entire series we have

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

    Substituting, we have

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

    Then we divide by $\displaystyle p+1$:

    $\displaystyle n^p\leq\frac{(n+1)^{p+1} - n^{p+1}}{p+1}\leq(n+1)^p$. $\displaystyle \blacksquare$
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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
    $\displaystyle
    n^p\leq(n+1)^{p-i}n^i\leq(n+1)^p
    $
    so this one is supposed to be:
    $\displaystyle
    n^p<(n+1)^{p-i}n^i<(n+1)^p
    $
    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:
    $\displaystyle
    n^p<(n+1)^{p-i}n^i<(n+1)^p
    $
    wich does not hold for i=0.

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