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Math Help - two more limits

  1. #1
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    two more limits

    Could someone show me these?

    Show that if lim as n goes to infinity (an) = A, then lim as n goes to infinity of
    na1 + (n-1)a2 + ... + 2an-1 + an
    ----------------------------- = (A/2)
    n^2


    and


    Show that for p in the natural numbers, the lim as n goes to infinity of
    (1^p) + (2^p) + ... + (n^p)
    ------------------------- = 1/(p+1)
    n^(p+1)

    Thanks in advance for any suggestions or help.
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  2. #2
    MHF Contributor red_dog's Avatar
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    All these limits you posted here are of the same kind and all can be solved by using the Stolz-Cesaro theorem. I don't know why your teacher gave you these problems if he didn't show you this theorem.

    Now, for the first one.
    Let x_n=na_1+(n-1)a_2+\ldots+2a_{n-1}+a_n, \ y_n=n^2.
    (y_n) is a positive sequence, strictly ascending and unbounded.

    \displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=
    \displaystyle=\lim_{n\to\infty}\frac{a_1+a_2+\ldot  s+a_{n+1}}{2n+1}=
    \displaystyle=\lim_{n\to\infty}\frac{a_1+a_2+\ldot  s+a_{n+1}}{n+1}\cdot\frac{n+1}{2n+1}=A\cdot\frac{1  }{2}=\frac{A}{2}.

    Here i used the following result:
    If \lim_{n\to\infty}a_n=A then \displaystyle\lim_{n\to\infty}\frac{a_1+a_2+\ldots  +a_n}{n}=A

    Then \displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\fra  c{A}{2}.


    The second one:
    Let x_n=1^p+2^p+\ldots+n^p, \ y_n=n^{p+1}.
    (y_n) is a positive sequence, strictly ascending and unbounded.

    \displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=
    \displaystyle=\lim_{n\to\infty}\frac{(n+1)^p}{(n+1  )^{p+1}-n^{p+1}}=\lim_{n\to\infty}\frac{n^p+C_p^1n^{p-1}+C_p^2n^{p-2}+\ldots+1}{n^{p+1}+C_{p+1}^1n^p+C_{p+1}^2n^{p-1}+\ldots+1-n^{p+1}}=
    \displaystyle=\lim_{n\to\infty}\frac{n^p+\ldots+1}  {(p+1)n^p+\ldots+1}=\frac{1}{p+1}.

    So, \displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\fra  c{1}{p+1}.
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  3. #3
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    Thanks a lot for helping.

    Um, this one is quite similar, but I can't seem to simplify it to the correct limit. Bare with me as I type this out.
    lim as n goes to infinity of (an) = A
    Find the limit as n goes to infinity of
    [(n)*a1 +
    (1)

    (n)*a2 + ........ +
    (2)

    (n)*an ]
    (n)
    --------------------------
    2^(n).

    where
    (n) = n!/[(n!)(n-n)!] (i.e. a combination without repetition)
    (n)
    The limit equals A.
    It seems I can't use the theorem because
    (n)
    (n+1)
    does not exist.

    I tried something else, but I'm getting tired of writing like this. Could you show me how to do this one as well?

    I really appreciate you helping.

    P.S. I'm sorry about the typing. I need to learn latex.
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  4. #4
    MHF Contributor red_dog's Avatar
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    \displaystyle\lim_{n\to\infty}\frac{C_n^1a_1+C_n^2  a_2+\ldots+C_n^na_n}{2^n}=
    \displaystyle=\lim_{n\to\infty}\frac{C_{n+1}^1a_1+  C_{n+2}^2a_2+\ldots+C_{n+1}^na_n+C_{n+1}^{n+1}a_{n  +1}-C_n^1a_1-C_n^2a_2-\ldots-C_n^na_n}{2^{n+1}-2^n}=
    \displaystyle=\lim_{n\to\infty}\frac{a_1(C_{n+1}^1-C_n^1)+a_2(C_{n+1}^2-C_n^2)+\ldots+a_n(C_{n+1}^n-C_n^n)+a_{n+1}}{2^n}=
    \displaystyle=\lim_{n\to\infty}\frac{C_n^0a_1+C_n^  1a_2+\ldots+C_n^{n-1}a_n+C_n^na_{n+1}}{C_n^0+C_n^1+\ldots+C_n^n}=A

    In the last limit, the fraction represents the arithmetic mean of the numbers
    a_1,\underbrace{a_2, a_2,\ldots,a_2}_{C_n^1},\underbrace{a_3,a_3,\ldots  ,a_3}_{C_n^2},\ldots,\underbrace{a_{n+1},a_{n+1},\  ldots,a_{n+1}}_{C_n^n}
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