Interval of Convergence

In summary: 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?
  • #1
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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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  • #2
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
Dick said:
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
hpayandah said:
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
Dick said:
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
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
hpayandah said:
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
Dick said:
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
 

1. What is an interval of convergence?

An interval of convergence is a range of values for which a given mathematical series will converge, meaning that the sum of the terms in the series will approach a finite value as the number of terms increases.

2. How do you determine the interval of convergence for a series?

To determine the interval of convergence for a series, you can use several methods, such as the Ratio Test, Root Test, or Alternating Series Test. These tests involve evaluating the limit of a sequence or a series and comparing it to known values to determine if the series converges or diverges.

3. Can a series have multiple intervals of convergence?

Yes, a series can have multiple intervals of convergence. This occurs when the series has different convergence behaviors at different points in the interval. For example, a power series may have an interval of convergence for which it converges absolutely and another interval for which it converges conditionally.

4. What happens if the value falls outside of the interval of convergence?

If a given value falls outside of the interval of convergence, the series will diverge, meaning that the sum of the terms will approach infinity rather than a finite value. This indicates that the series is not a valid representation of the function at that point.

5. Can the interval of convergence change for a given series?

Yes, the interval of convergence can change for a given series. This can occur if the series is modified or manipulated in some way, such as by taking a derivative or anti-derivative. These changes can affect the convergence behavior of the series and therefore alter the interval of convergence.

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