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Math Help - Find the Limit: Urgent. (1^p+2^p+.....+n^p) /(n^(p +1))

  1. #1
    sangal_ak04
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    Question Find the Limit: Urgent. (1^p+2^p+.....+n^p) /(n^(p +1))

    Hi All,

    I'm unable to find the exact steps to find out the answer of the following q?

    (1^p+2^p+.....+n^p)
    Limit ---------------------
    n->~(infinity) (n^(p+1))


    Thanks in advance.
    Arun Sangal
    91-9811514096
    arun.sangal@galileo.com
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  2. #2
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    Quote Originally Posted by sangal_ak04 View Post
    Hi All,

    I'm unable to find the exact steps to find out the answer of the following q?

    (1^p+2^p+.....+n^p)
    Limit ---------------------
    n->~(infinity) (n^(p+1))


    Thanks in advance.
    Arun Sangal
    91-9811514096
    arun.sangal@galileo.com
    There is a general formula to add,
    1^p+2^p+...+n^p = A_{p+1}n^{p+1}+A_pn^p+...+A_1+A_0
    The coefficients follow Bernoulli numbers and get messy.
    However the first ceofficient is simply A_{p+1}=\frac{1}{(p+1)}.

    So we have,
    \frac{\frac{1}{(p+1)}n^{p+1}+A_pn^p+...+A_1p+A_0}{  n^{p+1}}.
    And this limit is therefore,
    \boxed{ \frac{1}{(p+1)} }.
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  3. #3
    MHF Contributor red_dog's Avatar
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    Use the Stolz-Cesaro criteria.
    Let a_n=1^p+2^p+\ldots+n^p, \ b_n=n^{p+1}.
    (b_n)_{n\in\mathbf{N}} is ascendind and unbounded.
    We calculate \displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{(n+1)^p}{(n+1)^{p+1}-n^{p+1}}=
    \displaystyle =\lim_{n\to\infty}\frac{n^p+C_n^1n^{p-1}+\ldots}{n^{p+1}+C_{p+1}^1n^p+\ldots -n^{p+1}}=\frac{1}{p+1}

    So \displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=\fra  c{1}{p+1}.
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