anonymous
  • anonymous
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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jtvatsim
  • jtvatsim
Sorry, don't know what happened, there... it wasn't updating me that you had replied.
jtvatsim
  • jtvatsim
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.
jtvatsim
  • jtvatsim
\[\lim_{n \rightarrow \infty} \frac{11(n+7)^6}{10 (n+8)^6}\]

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jtvatsim
  • jtvatsim
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 \[\lim_{n \rightarrow \infty} \frac{11 \cdot (n^6 + \cdots)}{10 \cdot (n^6 + \cdots)}\]
jtvatsim
  • jtvatsim
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. \[\frac{11}{10}\lim_{n \rightarrow \infty} \frac{n^6 + \cdots}{n^6 + \cdots}=\frac{11}{10} \cdot 1 = \frac{11}{10}\]
jtvatsim
  • jtvatsim
This means that L > 1, and the series diverges.
anonymous
  • anonymous
great thanks so much!!! your awesome

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