Interval of Convergence

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Homework Statement


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


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.
 

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Answers and Replies

  • #2
Dick
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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.
 
  • #3
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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.
 
  • #4
Dick
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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.
 
  • #5
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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.
 

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  • #6
HallsofIvy
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It looks more complicated than necessary- it should be obvious that
[tex]\frac{3^{n+1}- 4}{3^n- 4}[/tex]
is dominated by [itex]3^{n+1}{3^n}= 3[/itex] and so its limit is 3. But your result, that the limit is [itex](3/4)|x- 1|[/itex] is correct. Now, what is the answer to your original question?
 
  • #7
Dick
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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.
 
  • #8
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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
 

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