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anonymous
 one year ago
could you also help me with the series (1)^(n+1)(11^(n1))/((n+7)^6(10^(n+2)) for ratio test to find L ?
anonymous
 one year ago
could you also help me with the series (1)^(n+1)(11^(n1))/((n+7)^6(10^(n+2)) for ratio test to find L ?

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jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, don't know what happened, there... it wasn't updating me that you had replied.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0So, 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \frac{11(n+7)^6}{10 (n+8)^6}\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0This means that L > 1, and the series diverges.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0great thanks so much!!! your awesome
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