What is the interval of convergance and how do you find it?

Question

I have a question that looks like this:
Which of the following is the interval of convergence of the power series:

the summation from n=0 to infinity of:(x-2)^n / 2^n

There are 5 different answers:
a) 2 <= x <2
b) -2 < x <= 2
c) 0 < x < 4
d) 0 <=x < 4
e) -2 <= x <= 2

Are there any tricks to solving these types of calculus problems? Do you know the answer to this one, and if so, how did you do it? I read about this absolute value ratio method, but cannot figure it out. Thanks for your help!

Answer 1

Look at the ratio of successive terms and see where this ratio has absolute value <1:

|((x-2)^(n+1) / 2^(n+1))/((x-2)^n / 2^n)|<1
if and only if
|(x-2) / 2|<1
if and only if
|x-2|<2
if and only if
0 This is the interval of convergence.

Answer 2

Hello,

The value at n=0 is 1 since (x-2)^0/2^0 = 1/1 and the value as x increases without bound approaches, but can never be 1 so the sum must be a.

Hope Tis Helps!

Answer 3

sum (x-2)^n/ 2^n = sum [ (x-2)/2 ]^n
this converges iff -1 < (x-2)/2 < 1
iff 0 < x/2 < 2
iff 0 so interval of convergence is 0 < x < 4 or (0,4)

(my iff means if and only if)