## 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 ?

1. jtvatsim

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

2. 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.

3. jtvatsim

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

4. 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 $\lim_{n \rightarrow \infty} \frac{11 \cdot (n^6 + \cdots)}{10 \cdot (n^6 + \cdots)}$

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

6. jtvatsim

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

7. anonymous

great thanks so much!!! your awesome