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Math Help - Limits of Sequences

  1. #1
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    Limits of Sequences

    Hey guys,
    I am struggling with the following homework problems:

    use the appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges.


    an = [en + (-3)n] / 5n [solved]

    an = en / 2n [solved]

    an = [1+(1/n)]n

    an = n[(n2 + 1)1/2 - n]

    Any help would be greatly appreciated
    Last edited by Calctaker; July 12th 2012 at 07:11 PM.
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  2. #2
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    Re: Limits of Sequences

    The first one converges. Realize that a_n = \frac{e^n}{5^n} + \frac{(-3)^n}{5^n} = (\frac{e}{5})^n + (-\frac{3}{5})^n, both sequences converge to zero.
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  3. #3
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    Re: Limits of Sequences

    Quote Originally Posted by Calctaker View Post
    an = [1+(1/n)]n
    \lim_{n\to\infty}\left(1+\frac1n\right)^n = e

    This is one of the basic representations of e.

    Quote Originally Posted by Calctaker View Post
    an = n[(n2 + 1)1/2 - n]
    \lim_{n\to\infty}n\left(\sqrt{n^2+1}-n\right)

    =\lim_{n\to\infty}\frac{n\left(\sqrt{n^2+1}-n\right)\cdot\left(\sqrt{n^2+1}+n\right)}{\sqrt{n^  2+1}+n}

    =\lim_{n\to\infty}\frac{n\left[\left(n^2+1\right)-n^2\right]}{\sqrt{n^2+1}+n}

    =\lim_{n\to\infty}\frac n{\sqrt{n^2+1}+n}

    =\lim_{n\to\infty}\frac{n/n}{\left(\sqrt{n^2+1}+n\right)/n}

    =\lim_{n\to\infty}\left(\sqrt{\frac{n^2+1}{n^2}}+1  \right)^{-1}

    =\lim_{n\to\infty}\left(\sqrt{1+\frac1{n^2}}+1 \right)^{-1}

    =\frac12
    Last edited by Reckoner; July 12th 2012 at 08:10 PM.
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    Re: Limits of Sequences

    Thank you very very much!!!!!!
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