# Interval of Convergence

## Homework Statement

Find the interval of convergence of each of the following
Ʃ$^{∞}_{n=0}$ ($\frac{3^{n}-2^{2}}{2^{2n}}$(x-1)$^{n}$)

## The Attempt at a Solution

Please refer to attachment. All I want to know is that I'm doing this problem right. I have found the interval but haven't plugged the interval back into the equation.

#### Attachments

• 20111125_002.jpg
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Dick
Homework Helper
You shouldn't break it into two series like that unless you know what you are doing. And combining them into (3/4)(x-1)-(1/4)(x-1) shows you probably don't. Just apply the ratio test to the whole expression.

You shouldn't break it into two series like that unless you know what you are doing. And combining them into (3/4)(x-1)-(1/4)(x-1) shows you probably don't. Just apply the ratio test to the whole expression.

The thing is I don't know what to do with that 2^2 in the equation, it throws me off. I'll re-post a picture this time everything together.

Dick
Homework Helper
The thing is I don't know what to do with that 2^2 in the equation, it throws me off. I'll re-post a picture this time everything together.

One of the factors in your ratio test should be (3^(n+1)-2^2)/(3^n-2^2), right? To find the limit of that as n->infinity, just divide numerator and denominator by 3^n.

One of the factors in your ratio test should be (3^(n+1)-2^2)/(3^n-2^2), right? To find the limit of that as n->infinity, just divide numerator and denominator by 3^n.

I tried your suggestion, however I didn't come to a good end (maybe I did something wrong). Please take a look at the attached file; this is my re-attempt.

#### Attachments

HallsofIvy
Homework Helper
It looks more complicated than necessary- it should be obvious that
$$\frac{3^{n+1}- 4}{3^n- 4}$$
is dominated by $3^{n+1}{3^n}= 3$ and so its limit is 3. But your result, that the limit is $(3/4)|x- 1|$ is correct. Now, what is the answer to your original question?

Dick
Homework Helper
I tried your suggestion, however I didn't come to a good end (maybe I did something wrong). Please take a look at the attached file; this is my re-attempt.

Not right. There's a mistake in the long division. 3^(n+1)-3^n isn't 3. And I didn't mean that in my suggestion. Take (3^(n+1)-4)/(3^n-4). Dividing the numerator, 3^(n+1)-4 by 3^n gives you 3^(n+1)/3^n-4/3^n. What's 3^(n+1)/3^n? Now take the limit as n->infinity. Do the same with the denominator.

Not right. There's a mistake in the long division. 3^(n+1)-3^n isn't 3. And I didn't mean that in my suggestion. Take (3^(n+1)-4)/(3^n-4). Dividing the numerator, 3^(n+1)-4 by 3^n gives you 3^(n+1)/3^n-4/3^n. What's 3^(n+1)/3^n? Now take the limit as n->infinity. Do the same with the denominator.
Thank you I got it now.

It looks more complicated than necessary- it should be obvious that
3n+1−43n−4

is dominated by 3n+13n=3 and so its limit is 3. But your result, that the limit is (3/4)|x−1| is correct. Now, what is the answer to your original question?
Thanks