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Thread: Brief limit questions.

  1. #1
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    Brief limit questions.

    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    $\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

    {$\displaystyle s_{n}$} is increasing.

    Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?
    Last edited by CountingPenguins; Jun 29th 2011 at 09:49 PM. Reason: Figured out the formatting.
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  2. #2
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    Re: Brief limit questions.

    You need to use tex tags, not math. And your close tag needs /, not \.
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  3. #3
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    Re: Brief limit questions.

    Thanks for the pointer on the formatting. And I think the Lady Gaga equation at the bottom is HILARIOUS!
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    Re: Brief limit questions.

    Quote Originally Posted by CountingPenguins View Post
    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    $\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

    {$\displaystyle s_{n}$} is increasing.

    Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?
    $\displaystyle s_1=1, s_{n+1}=\frac{1}{4}(s_n+5)$

    $\displaystyle s_n$ is monotone increasing:

    Induction on n:

    Say holds for $\displaystyle n$: $\displaystyle s_{n+1}>s_n$

    Proving for $\displaystyle n+1$:

    $\displaystyle s_{n+2}>s_{n+1}$

    $\displaystyle s_{n+2}=\frac{1}{4}(s_{n+1}+5)\overset{?}{>}s_{n+1 }=\frac{1}{4}(s_n+5)$

    .
    .
    .
    $\displaystyle s_{n+1}>s_n$



    $\displaystyle s_n$ is bounded:

    Induction on $\displaystyle n$:

    Say that $\displaystyle s_n<2$ for all $\displaystyle n$ is true.

    Proving for $\displaystyle n+1$:

    $\displaystyle s_{n+1}=\frac{1}{4}(s_n+5)<\frac{2}{4}+\frac{5}{4} =\frac{7}{4}<2$

    So, $\displaystyle s_n$ is monotonic increasing and bounded, hence $\displaystyle \lim_{n\to\infty}s_n$ exists.

    We denote $\displaystyle \lim_{n\to\infty}s_n=l$ where $\displaystyle l\in\mathbb{R}$.

    $\displaystyle \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}\frac{1} {4}(s_n+5)$

    $\displaystyle \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}s_n=l$

    $\displaystyle l=\frac{1}{4}(l+5)$

    $\displaystyle l=\frac{5}{3}$
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Brief limit questions.

    Quote Originally Posted by CountingPenguins View Post
    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    $\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

    {$\displaystyle s_{n}$} is increasing.

    Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?
    Another solution:



    $\displaystyle s_{n+1}=\frac{1}{4}s_n+\frac{5}{4}$, $\displaystyle s_1=1$

    Define new sequence: $\displaystyle b_n$ :

    $\displaystyle b_n:=s_n-\frac{5}{3}$

    $\displaystyle b_n$ is geometric sequence with $\displaystyle q=\frac{1}{4}$ and $\displaystyle b_1=\frac{-2}{3}$.

    Hence, $\displaystyle b_n=\frac{-2}{3}\frac{1}{4^{n-1}}$.

    So, $\displaystyle s_n=b_n+\frac{5}{3}=\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}$

    $\displaystyle s_n=\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}$

    $\displaystyle \lim_{n\to\infty}s_n=\lim_{n\to\infty}\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}=$

    $\displaystyle =0+\frac{5}{3}=\frac{5}{3}$
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    Re: Brief limit questions.

    Quote Originally Posted by CountingPenguins View Post
    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    $\displaystyle s_{1}$=1 and $\displaystyle s_{n+1}$=$\displaystyle \frac{1}{4}$($\displaystyle s_{n}$+5)for n $\displaystyle \in$ N.

    {$\displaystyle s_{n}$} is increasing.

    Are we guessing that $\displaystyle s_{n}$ $\displaystyle \leq 2$?
    First observe that if this converges it converges to a root of:

    $\displaystyle s=\frac{1}{4}(s+5)$

    knowing this greatly simplifies your work (in this case it provides a better guess for the upper bound for one thing).

    CB
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