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Math Help - converging/diverging of geometric series

  1. #1
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    Unhappy converging/diverging of geometric series

    I have a question about the lim as it approaches infinity of ((-6)^(n-1))/(5^(n-1)). I am unsure what to do.. completely actually. I was thinking i would try to put it in a/(1-r) mode, but i dont really know how, or if that is even the right thing to do.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RhotelidlewildR View Post
    I have a question about the lim as it approaches infinity of ((-6)^(n-1))/(5^(n-1)). I am unsure what to do.. completely actually. I was thinking i would try to put it in a/(1-r) mode, but i dont really know how, or if that is even the right thing to do.
    You said series. So I assume you are talking about \sum_{n=1}^{\infty}\frac{(-6)^{n-1}}{5^{n-1}}. Merely note that \left|\frac{(-6)^{n-1}}{5^{n-1}}\right|\ge1...so what?
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  3. #3
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    Smile finding limits.

    I need to find whether that is converging or diverging, and if convergence i need to find the limit. But yes, that is what i meant.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RhotelidlewildR View Post
    I need to find whether that is converging or diverging, and if convergence i need to find the limit. But yes, that is what i meant.
    But my dear RhotelidlewildR I did tell you whether or not it converges.
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  5. #5
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    thanks.

    Quote Originally Posted by Drexel28 View Post
    But my dear RhotelidlewildR I did tell you whether or not it converges.
    yeah, so unless i am wrong (which i very well could be) it is greater than one so it converges. But my problem is breaking it down to what value it converges at. Do i divide by ln(n) since it is the loswest power in the numerator? I am just really lost. Thanks for your patience!
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by RhotelidlewildR View Post
    yeah, so unless i am wrong (which i very well could be) it is greater than one so it converges. But my problem is breaking it down to what value it converges at. Do i divide by ln(n) since it is the loswest power in the numerator? I am just really lost. Thanks for your patience!
    Now I am really lost. Where did this \ln(n) come from? Also, if the series is indeed what I posted in my seminal post then the series diverges.
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  7. #7
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    haha

    i am sorry i accidentally ran off on another problem. So is this the way i do it... a=-1, r=(6/5), but if that is the case i end up with -1/-(1/5) or 5. this is confusing the life out of me.
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  8. #8
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    If  \lim_{n \rightarrow \infty} \frac{(-6)^{n-1}}{5^{n-1}} \neq 0 then  \sum_{i = 1}^{\infty} \frac{(-6)^{n-1}}{5^{n-1}} diverges.
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