Question

Find the interval of convergence (I) for the series:
n approaches infintiy (x-2)^(2n) / 9^n

Answer 1

Use ratio test to find a condition on x.

Answer 2

Do you know how to use the nth-root test or ratio test?

Answer 3

yes. i used ratio test. i found that the radius of convergence is 5. is that correct?

Answer 4

now i'm geting the radius of convergence to be 1

Answer 5

The interval of convergence should be from (-1, 5) but you need to test the endpoints to see if they in fact converge or not.

Answer 6

yeah, can you explain to me how you got there. Using ratio test I got the limit as n approaches infinity of (x-2)^2 /9 Then I used the inequality (x-2)^2 /9 <1

Answer 7

so I did (x-2)^2 < 9 and then i square rooted each side and got |x-2| < 3

Answer 8

then i added 2 and got |x| < 5 . so my interval was (-5, 5) how did you get (-1,5)

Answer 9

Not quite, but almost. When |x-2|<3, that should be rewritten into the following inequality: -3< x-2 < 3 because you have two cases to consider. What's inside can either be positive, or what's inside can be negative. If you solve this inequality, you will find the interval goes from -1 < x < 5 and then check your endpoints.

Answer 10

oh yeah! I totally forgot. thank you!

Answer 11

NP! Glad to help.