Question

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 ?

Answer 1

Sorry, don't know what happened, there... it wasn't updating me that you had replied.

Answer 2

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.

Answer 3

\[\lim_{n \rightarrow \infty} \frac{11(n+7)^6}{10 (n+8)^6}\]

Answer 4

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
\[\lim_{n \rightarrow \infty} \frac{11 \cdot (n^6 + \cdots)}{10 \cdot (n^6 + \cdots)}\]

Answer 5

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.
\[\frac{11}{10}\lim_{n \rightarrow \infty} \frac{n^6 + \cdots}{n^6 + \cdots}=\frac{11}{10} \cdot 1 = \frac{11}{10}\]

Answer 6

This means that L > 1, and the series diverges.

Answer 7

great thanks so much!!! your awesome