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Math Help - Limit Comparison Test

  1. #1
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    Limit Comparison Test

    \sum^ \infty _{n=1} \frac{2}{3^n-5} = \lim_{n\rightarrow \infty} \frac{6^n}{6^n-25} = 1 > 0

    This is what I think:

    The original function is divergent since \sum^ \infty _{n=1} \frac{2}{3^n} is divergent.

    Is this right? Could you explain to me how to tell if the function is div. or conv. based on my Comparison Test???
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by saiyanmx89 View Post
    \sum^ \infty _{n=1} \frac{2}{3^n-5} = \lim_{n\rightarrow \infty} \frac{6^n}{6^n-25} = 1 > 0

    This is what I think:

    The original function is divergent since \sum^ \infty _{n=1} \frac{2}{3^n} is divergent.

    Is this right? Could you explain to me how to tell if the function is div. or conv. based on my Comparison Test???
    This series is convergent

    \sum^ \infty _{n=1} \frac{2}{3^n}=2\cdot \frac{\frac{1}{3}}{1-\frac{1}{3}}=1

    It is a geometric series with r=\frac{1}{3}

    How will this change your conclusion
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