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Math Help - Sequences: why does this converge?

  1. #1
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    Sequences: why does this converge?

    Just started going over sequences, and was wondering why how these 2 problems are convergent. Thanks.

    1.)

    solution posted:



    2.)

    solution posted:
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  2. #2
    Senior Member yeKciM's Avatar
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    1)

    a_n=\frac{2^n}{3^{n+1}}

    a_{n+1}=\frac{2^{n+1}}{3^{n+2}}

    so u have

    \displaystyle \lim_{n\to\infty} \frac {a_{n+1}}{a_n}= \lim_{n\to\infty} \frac {\frac{2^{n+1}}{3^{n+2}}}{\frac{2^n}{3^{n+1}}}= \lim_{n\to\infty} \frac {\frac{2^n\cdot2}{3^{n+1}\cdot3}}{\frac{2^n}{3^{n+  1}}}=\lim_{n\to\infty}\frac {2}{3}= \frac {2}{3}<1

    so it converge



    for the second I don't see where u have any problems
    same thing do as in first (maybe u have problems with " !") but i think it's not that, because you're doing sequences, and factorials must be done way more before sequences

    Edit : this is wrong sorry
    Last edited by yeKciM; August 5th 2010 at 07:14 AM.
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  3. #3
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    Re the first reply.
    This question is about sequences and is not about series.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Plato View Post
    Re the first reply.
    This question is about sequences and is not about series.
    i'm so sorry i just got a little a bit confused , while translating it's just the same sorry and i forgot what is which and didn't quite look at it sorry again
    Last edited by yeKciM; August 5th 2010 at 07:12 AM.
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  5. #5
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    Quote Originally Posted by Evan.Kimia View Post
    Just started going over sequences, and was wondering why how these 2 problems are convergent. Thanks.

    1.)

    solution posted:



    2.)

    solution posted:
    \displaystyle\lim_{n\rightarrow\infty}\left(\frac{  2}{3}\right)^n=\lim_{n\rightarrow\infty}\left(\fra  c{\left[\frac{2}{2}\right]}{\left[\frac{3}{2}\right]}\right)^n=\lim_{n\rightarrow\infty}\frac{1}{\left  (\frac{3}{2}\right)^n}

    As n\rightarrow\infty the denominator keeps increasing.

    \frac{1}{infinitely\ large}=infinitely\ small

    The second one has an infinitely large value in the denominator position as n increases without limit.
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  6. #6
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    I pasted this same reply at the wrong place and got bit by the lion in the den. Here I am in the right place and doing it once again, just for you..

    I can’t see what you posted, but if you want to know why the sequence a_n=\frac{2^n}{3^{n+1}} converges, you can imagine 1 remains stationary, while the 3^n in the denominator increases at a much faster pace than the numerator 2^n, and eventually it converges to 0.
    Once we know that it converges to 0, we can prove that with a given \epsilon>0, there is a positive integer N such that if |\frac{2^n}{3^{n+1}}-0|=|\frac{2^n}{3^{n+1}}|<\epsilon, but we know that |\frac{2^n}{3^{n+1}}<|\frac{2^n}{3^n}|<\epsilon, so we can use |\frac{2^n}{3^n}|=|(\frac{2}{3})^n|=|(\frac{1}{(3/2)^n})|<\epsilon to find n.

    Since (3/2)^n> \frac{1}{\epsilon}, which is equivalent to n \log 1.5>\log \frac{1}{\epsilon}, which also is equivalent to n>\log_{1.5} (\frac{1}{\epsilon}), we are now ready to prove.

    Proof:
    Let  \epsilon >0. Choose  N=max(1,\lceil \log_{1.5} \frac{1}{\epsilon}\rceil ) and let n be any integer such that  n>N . Thus n>\log_{1.5} (\frac{1}{\epsilon}), and so |\frac{2^n}{3^{n+1}}-0|=|\frac{2^n}{3^{n}}|<|\frac{1}{(3/2)^n}<\epsilon .
    Q.E.D.
    Last edited by novice; August 5th 2010 at 03:38 PM. Reason: mistake
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