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anonymous

  • one year ago

could you also help me with the series (-1)^(n+1)(11^(n-1))/((n+7)^6(10^(n+2)) for ratio test to find L ?

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  1. jtvatsim
    • one year ago
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    Sorry, don't know what happened, there... it wasn't updating me that you had replied.

  2. jtvatsim
    • one year ago
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    So, we can throw away the negative sign since we have absolute values, but it still looks hard to evaluate the limit of the n's since we have powers.

  3. jtvatsim
    • one year ago
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    \[\lim_{n \rightarrow \infty} \frac{11(n+7)^6}{10 (n+8)^6}\]

  4. jtvatsim
    • one year ago
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    If we are clever, we will realize though that IF we expanded those sixth powers, the highest power would be a n^6 in both the top and bottom \[\lim_{n \rightarrow \infty} \frac{11 \cdot (n^6 + \cdots)}{10 \cdot (n^6 + \cdots)}\]

  5. jtvatsim
    • one year ago
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    The other powers are irrelevant as we take the limit to infinity, leading us to the conclusion that the n^6 powers will approach 1 in the limit. \[\frac{11}{10}\lim_{n \rightarrow \infty} \frac{n^6 + \cdots}{n^6 + \cdots}=\frac{11}{10} \cdot 1 = \frac{11}{10}\]

  6. jtvatsim
    • one year ago
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    This means that L > 1, and the series diverges.

  7. anonymous
    • one year ago
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    great thanks so much!!! your awesome

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