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Math Help - series and sequence help

  1. #1
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    series and sequence help

    sorry for the image but i am having trouble proving the answers i get while doing my first problem



    (a) I know the sequence converges to 0 since if you take the limit as n goes to infinity, then a(n) will go to 0 because fractions multiplied by themselves many times that are below 1 go to 0. but how do i prove this answer??

    (b) is this geometric?

    (c) does it converge because its a constant ratio by geometric series test?

    (d) does it converge to 9?

    (e) i know that the answer is yes, .999... is equal to 1. but how do i prove that? would i set up a series of some kind??

    thanks
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  2. #2
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    Quote Originally Posted by Arez View Post

    (b) is this geometric?

    For your series \{t_1,t_2,t_3,\cdots\} = \{\left(\frac{9}{10}\right)^1,\left(\frac{9}{10}\r  ight)^2,\left(\frac{9}{10}\right)^3,\cdots\}

    It is geometric if \frac{t_3}{t_2} = \frac{t_2}{t_1}
    Last edited by pickslides; July 23rd 2009 at 05:51 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by Arez View Post

    (c) does it converge because its a constant ratio by geometric series test?

    Only if



    \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right  |<1
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  4. #4
    MHF Contributor matheagle's Avatar
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    (e) Let
    x=.\bar 9
    then
    10x=9.\bar 9
    subtract x from both sides, x from the left and .\bar 9 from the right

    so 9x=9 or x=1.
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  5. #5
    MHF Contributor matheagle's Avatar
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    (b) geometric with r=.9, Geo's converge when -1<r<1.
    The ratio test is basically a limit comparison to a geometric.
    The limit of adjacent terms is the search for the essential r.

    (c) and (d) yes it converges to 9...

    the limit is

    {9/10\over 1-9/10}={9\over 10-9}=9
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  6. #6
    MHF Contributor matheagle's Avatar
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    (a) (.9)^n\to 0 as n goes to infinity
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  7. #7
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    Yes, the original poster said he knew that in his original post.
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