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Math Help - Proof By Inducion.

  1. #1
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    Proof By Inducion.

    So I need to proof the following inequality

    \sum_{k\ =\ 0}^{n-1} k^{p} < \frac{n^{p+1}}{p+1} < \sum_{k\ =\ 0}^n k^{p}

    If i start it by induction, I get to a point where I have to proove that

    k^p (k+p+1) < (k+1)^{p+1} (with the first part)

    And I have no idea where to go from here. Any suggestions?
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by ml_lulu View Post
    So I need to proof the following inequality

    \sum_{k\ =\ 0}^{n-1} k^{p} < \frac{n^{p+1}}{p+1} < \sum_{k\ =\ 0}^n k^{p}

    If i start it by induction, I get to a point where I have to proove that

    k^p (k+p+1) < (k+1)^{p+1} (with the first part)

    And I have no idea where to go from here. Any suggestions?
    (k+1)^{p+1}=k^{p+1}+{\color{red}C_{p+1}^p k^p}+{\color{blue}C_{p+1}^{p-1} k^{p-1}+\dots+C_{p+1}^1 k+1}

    {\color{red}C_{p+1}^p}=\frac{(p+1)!}{p!(p+1-p)!}={\color{red}p+1}

    All the blue part is positive (and different to 0 unless k is somehow negative), do you agree ?

    Therefore, we get :

    (k+1)^{p+1}={\color{magenta}k^{p+1}+(p+1)k^p}+{\co  lor{blue}\text{blue thing}} > {\color{magenta} k^{p+1}+(p+1)k^p}=k^p(k+p+1) \ \square

    Does it make sense ?
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi

    Another possible approach, without induction :

    Let k\in\mathbb{N}.

    Using the properties of the integral, \int_k^{k+1} t^p\,\mathrm{d}t \geq \int_k^{k+1} k^p\,\mathrm{d}t=(k+1-k)k^p=k^p

    Summing these inequalities for k\in\{0,\,1\ldots,\,n-1\} gives \sum_{k=0}^{n-1} k^p\leq \sum_{k=0}^{n-1} \int_k^{k+1} k^p\,\mathrm{d}t

    The right hand side can be rewritten as \sum_{k=0}^{n-1} \int_k^{k+1} k^p\,\mathrm{d}t=\int_0^{1} k^p\,\mathrm{d}t+\int_1^{2} k^p\,\mathrm{d}t+\int_2^{3} k^p\,\mathrm{d}t+\ldots+\int_{n-1}^{n} k^p\,\mathrm{d}t=\int_0^n k^p\,\mathrm{d}t hence \sum_{k=0}^{n-1} k^p\leq\int_0^n k^p\,\mathrm{d}t .

    As \int_0^{n} k^p\,\mathrm{d}t=\left[ \frac{t^{p+1}}{p+1}\right]_0^n=\frac{n^{p+1}}{p+1}, here is the first inequality : \boxed{\sum_{k=0}^{n-1} k^p\leq\frac{n^{p+1}}{p+1}}.

    Using the same method, one can show the other one.
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