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Math Help - Limits of sequences

  1. #1
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    Limits of sequences

    Describe the limiting behaviour of the following sequences:

    1) \dfrac{\ln n}{n^a}, a>0

    I think its 0. I worked this out because I know n^a grows faster than \ln n. Is that good enough to do that in a exam situation, or would you need to prove it.



    2) \dfrac{n\sin (n\pi/4)}{\sqrt{n^2+1}}

    I think this diverges, but I'm not too sure how to show it.



    3) \dfrac{n+\cos n\pi}{n-\cos n\pi}

    Would I divide top and bottom by n and so the sequence would converge to -1
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  2. #2
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    When you say "the limiting behaviour", I assume you mean the limit as n \to \infty.

    For your first:

    \lim_{n \to \infty}\frac{\ln{n}}{n^a} tends to \frac{\infty}{\infty}, so you can use L'Hospiatal's Rule.

    \lim_{n \to \infty}\frac{\ln{n}}{n^a} = \lim_{n \to \infty}\frac{\frac{1}{n}}{an^{a - 1}}

     = \lim_{n \to \infty}\frac{1}{an^a}

     = 0.


    2. Note that \sin{x} \leq 1.

    Therefore \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^  2 + 1}} \leq \frac{n}{\sqrt{n^2 + 1}}.

    So if the second sequence is convergent, so is the first.

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

     = 1.

    Since \frac{n}{\sqrt{n^2 + 1}} is convergent, so is \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^  2 + 1}}.


    3. \lim_{n \to \infty}\frac{n + \cos{(n\pi)}}{n - \cos{(n\pi)}} =\lim_{n \to \infty} \frac{n - \cos{(n\pi)} + 2\cos{(n\pi)}}{n - \cos{(n\pi)}}

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

     = 1 since in the fraction the top oscillates while the bottom grows without bound.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Prove It View Post
    2. Note that \sin{x} \leq 1.

    Therefore \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^  2 + 1}} \leq \frac{n}{\sqrt{n^2 + 1}}.

    So if the second sequence is convergent, so is the first.

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

     = 1.

    Since \frac{n}{\sqrt{n^2 + 1}} is convergent, so is \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^  2 + 1}}.
    Wrong. It does not converge since \sin(n\pi/4) never settles down this cycles through values close to 0, \sqrt{2}, 1, \sqrt{2}, 0,-\sqrt{2},-1,-\sqrt{2} .

    It's a sequence not a series.

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    Wrong. It does not converge since \sin(n\pi/4) never settles down this cycles through values close to 0, \sqrt{2}, 1, \sqrt{2}, 0,-\sqrt{2},-1,-\sqrt{2} .

    It's a sequence not a series.

    CB
    Yes, you are right, sorry. All I did was show that the values are always no greater than 1.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Prove It View Post
    3. \lim_{n \to \infty}\frac{n + \cos{(n\pi)}}{n - \cos{(n\pi)}} =\lim_{n \to \infty} \frac{n - \cos{(n\pi)} + 2\cos{(n\pi)}}{n - \cos{(n\pi)}}

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

     = 1 since in the fraction the top oscillates while the bottom grows without bound.
    Simpler (and closer to what the OP tried to do):

    \dfrac{n+\cos(n\pi)}{n-\cos(n\pi)}=\dfrac{1+\frac{\cos(n\pi)}{n}}{1-\frac{\cos(n\pi)}{n}}

    So as top and bottom of the right hand side both go to 1 as n\to \infty the limit of the whole thing is 1.

    CB
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