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Thread: Help with a limit

  1. #1
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    Help with a limit

    $\displaystyle \lim_{n \to \infty} (\frac{3}{4})^n$

    I know the answer is 0. But exactly how can I show that?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Cakecake View Post
    $\displaystyle \lim_{n \to \infty} (\frac{3}{4})^n$

    I know the answer is 0. But exactly how can I show that?
    By prove it what tools do you mean? By delta-epsilon defintion?

    We can prove that if $\displaystyle p>0,\alpha\in\mathbb{R}$ then $\displaystyle \lim_{n\to\infty}\frac{n^{\alpha}}{(1+p)^n}=0$

    Proof: Consider an integerl $\displaystyle K>\alpha,K>0$. For $\displaystyle n>2K$

    $\displaystyle (1+p)^n> {n\choose {K}} p^K>\frac{n^Kp^K}{2^KK!}$

    This implies that

    $\displaystyle 0\leqslant\frac{n^{\alpha}}{(1+p)^n}\leqslant\frac {2^KK!}{p^K}n^{\alpha-K}~~~n>2K$

    Now you should know that if $\displaystyle \beta>0$ that $\displaystyle \lim_{n\to\infty}\frac{1}{n^{\beta}}=0$...so since $\displaystyle \alpha-K<0$ this proves that $\displaystyle \lim_{n\to\infty}\frac{n^{\alpha}}{(1+p)^n}=0$...take $\displaystyle \alpha=0,p=\frac{1}{3}$ to acheive your limit.


    Is that what you wanted? Or were you looking for something simpler?


    Note: I think if you weren't looking for a proof like the one above you could just state that $\displaystyle \left|\frac{3}{4}\right|<1$ so $\displaystyle \lim_{n\to\infty}\left(\frac{3}{4}\right)=0$
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  3. #3
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    Woah I was looking for a much simpler answer.

    Thanks anyway for typing all that up.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Cakecake View Post
    Woah I was looking for a much simpler answer.

    Thanks anyway for typing all that up.
    Then I think it would suffice to say that since $\displaystyle \left|\frac{3}{4}\right|<1$ that $\displaystyle \lim_{n\to\infty}\left(\frac{3}{4}\right)^n=0$. What level are you currently at?
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  5. #5
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    I guess that could work.

    I'm taking BC, currently a junior. I'm up to series.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Cakecake View Post
    I guess that could work.

    I'm taking BC, currently a junior. I'm up to series.
    Perfect ! Consider that $\displaystyle \lim_{n\to\infty}\sqrt[n]{\left(\frac{3}{4}\right)^n}=\frac{3}{4}<1\implies \sum_n \left(\frac{3}{4}\right)^n\text{ converges}$. Now what is one of the neccessary conditions for series convergence?
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  7. #7
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    The limit of it as n goes to infinity has to be 0.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Cakecake View Post
    The limit of it as n goes to infinity has to be 0.
    ...well there you go
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