What Is the Interval of Convergence for the Series Summation?

[hpayandah]

Homework Statement

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

Homework Equations

Please refer to attachment

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.

 

[Dick]

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.

 

[hpayandah]

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]

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.

 

[hpayandah]

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.

 

[HallsofIvy]

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]

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.

 

[hpayandah]

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