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Math Help - Convergance

  1. #1
    ADY
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    How can i decide if these sequence converge, then go on to state its limit?

    a_n = \frac {9 + 8(0.6)^n}{6(0.3)^n - 6} (n=1,2,3,...)

    Thanks guys

    do i start like this:



    a_n = \frac {9 + (4.8)^n}{(1.8)^n - 6}
    Last edited by mr fantastic; November 15th 2009 at 04:30 AM. Reason: Merged posts
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  2. #2
    Flow Master
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    Quote Originally Posted by ADY View Post
    do i start like this:



    a_n = \frac {9 + (4.8)^n}{(1.8)^n - 6}
    You seem to have the misunderstanding that a (b)^n = (ab)^n.

    Both (0.6)^n and (0.3)^n approach zero as n approaches infinity. What does that tell you?
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  3. #3
    ADY
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    They are divergent?
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  4. #4
    Senior Member Sampras's Avatar
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    Quote Originally Posted by ADY View Post
    They are divergent?
    it tells you that  \lim\limits_{n \to \infty} a_n \neq 0 . And that implies......?
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  5. #5
    ADY
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    Quote Originally Posted by Sampras View Post
    it tells you that  \lim\limits_{n \to \infty} a_n \neq 0 . And that implies......?
    It is convergent?
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  6. #6
    Senior Member Sampras's Avatar
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    Quote Originally Posted by ADY View Post
    It is convergent?
    no, divergent.
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  7. #7
    ADY
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    Ok, thank you - i really need to understand this further. Ill try another example

    b_n = \frac{2 - 7n^2}{3n + 4}
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  8. #8
    ADY
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    so do we say

    b_n = \frac{2 - 7n^2}{3n + 4} \rightarrow b_n = \frac{2 - 7n}{3 + 4/n}

    so 3 + 4/n converges to 3, so it convergent? Whats it limit
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  9. #9
    Flow Master
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    Quote Originally Posted by ADY View Post
    How can i decide if these sequence converge, then go on to state its limit?

    a_n = \frac {9 + 8(0.6)^n}{6(0.3)^n - 6} (n=1,2,3,...)

    Thanks guys

    do i start like this:



    a_n = \frac {9 + (4.8)^n}{(1.8)^n - 6}
     \lim_{n \rightarrow + \infty} a_n = \lim_{n \rightarrow + \infty}\frac {9 + 8(0.6)^n}{6(0.3)^n - 6} = \frac{9 + 0}{0 - 6} = ....

    Therefore the sequence (NB: not series) converges.

    Quote Originally Posted by Sampras View Post
    it tells you that  \lim\limits_{n \to \infty} a_n \neq 0 . And that implies......?
    It is a sequence not a series.
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