Infinite Series -

Find the interval of convergence for the following geometric series and, within this interval, find the sum;

$$\sum_{n=2}^{Infinity} (x-1)^{n-1}/2^{n+1}$$
$$\sum_{n=0}^{Infinity} (x-1)^{n+1}/2^{n+3}$$
=(x-1)^n (x-1)^1 / (2^n * 2^3)
= (x-1)/8 * (x-1/2)^n

Sort of stuck where to go from here. I wasn't sure if A = (x-1)/8 and r= (x-1)/2 and I just use a/(1-r) or what. Any help would be appreciated!

Question

Infinite Series - 

Find the interval of convergence for the following geometric series and, within this interval, find the sum;

$$\sum_{n=2}^{Infinity} (x-1)^{n-1}/2^{n+1}$$
$$\sum_{n=0}^{Infinity} (x-1)^{n+1}/2^{n+3}$$
=(x-1)^n (x-1)^1 / (2^n * 2^3)
= (x-1)/8 * (x-1/2)^n

Sort of stuck where to go from here. I wasn't sure if A = (x-1)/8 and r= (x-1)/2 and I just use a/(1-r) or what. Any help would be appreciated!

amistre64 · Answer

are you trying to find where the 2 of them meet?

anonymous · Answer

Not entirely sure what that means.

anonymous · Answer

The sandwich method? I was using that for Sequences.

amistre64 · Answer

well, ive done intervals of convergence on a single summation; but ive never even considered doing it for 2 at the same time

amistre64 · Answer

i assume it wants the interval where they meet at

anonymous · Answer

Again, not sure what you mean by 2 at the same time. If you mean the first two equations, the first is N=2 and then I did N = 0 and added two to N.

anonymous · Answer

If you have another method in mind for solving it I would love to hear it. I think what is getting me is the "X". If it were a 2, instead of X, I could have 1/8 instead of (x-1)/8, which would be my A, and then (x-1)/2 would be my R and I would use a/(1-r)

amistre64 · Answer

$$\lim_{n\to inf}\frac{(x-1)^{n-1}}{2^{n+1}}\frac{2^{n}}{(x-1)^{n-2}}$$

$$\lim_{n\to inf}\frac{(x-1)}{2}=\frac{x-1}{2}$$

amistre64 · Answer

i thought you had to sums there to play with; but it looks like you just altered the second one from teh first

amistre64 · Answer

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

anonymous · Answer

Yeah - random question - how do you get the division to go one on top of the other?

anonymous · Answer

Thanks for the link, Ill check it out.

amistre64 · Answer

the limit is a ratio of $$\frac{a_n}{a_{n-1}}$$

and the fraction thing is the code:  \frac{top}{bottom}

amistre64 · Answer

if im reading the site right;  the interval of convergence is then:
$$\frac{|x-1|}{2}<1$$
$$|x-1|<2$$
-1 < x < 3

amistre64 · Answer

http://www.wolframalpha.com/input/?i=sum+%28x%E2%88%921%29%5E%28n%E2%88%921%29+%2F%282%5E%28n%2B1%29%29+from+0+to+inf++ and i think the wolf agrees ;)