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Thread: Limits of sequences

  1. #1
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    Limits of sequences

    Describe the limiting behaviour of the following sequences:

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

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



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

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



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

    Would I divide top and bottom by $\displaystyle n$ 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 $\displaystyle n \to \infty$.

    For your first:

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

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

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

    $\displaystyle = 0$.


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

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

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

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

    $\displaystyle = 1$.

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


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

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

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

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

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

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

    $\displaystyle = 1$.

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

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

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

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

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

    CB
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