A Sequence in a Power Series

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Summary:

{##A_n##} is a sequence of positive numbers. ##\sum_{n=0}^{\infty} A_n(x-1)^n## has a R of convergence ##R=\frac 3 2##. Does ##\sum_{n=0}^{\infty} A_n## converge or diverge?
I tried to use the ratio test, but I am stuck on finding the range of the limit.
$$\because \left|x-1\right|<1.5=Radius$$
$$\therefore -0.5<x<2.5$$

$$\lim _{n \to \infty} \left| \frac{A_{n+1}(x-1)^{n+1}}{A_n(x-1)^n} \right|$$
$$\lim_{n \to \infty} \frac{A_{n+1} \left|x-1\right|}{A_n} <1$$
$$\lim_{n \to \infty} \left| \frac{A_{n+1}}{A_n} \right|< \frac 1 {\left| x-1 \right|}$$

Then I plotted the graph of 1/|x-1| and I found that the limit of A_n+1/A_n could vary from 2/3 (convergent) to infinity (divergent), as shown by the image below. What should I do next?
1593692590251.png
 
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  • #2
PeroK
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What happens if you take ##x = 2##?
 
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  • #3
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What happens if you take ##x = 2##?
No conclusion?
 
  • #4
PeroK
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No conclusion?
What does that mean? Is ##2## inside or outside the radius?
 
  • #5
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What does that mean? Is ##2## inside or outside the radius?
Does this show the radius of convergence ##R'## of the series A_n is 2?
 
  • #6
PeroK
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It means we have to use another method to test the convergence of the series. 2 is in the radius R.
Eh?

Summary:: {##A_n##} is a sequence of positive numbers. ##\sum_{n=0}^{\infty} A_n(x-1)^n## has a R of convergence ##R=\frac 3 2##.
 
  • #7
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Eh?
I think it is in the radius of convergence because ##-0.5<2<2.5##
 
  • #8
PeroK
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I think it is in the radius of convergence.
You're saying that you think ##2 < \frac 3 2##?
 
  • #9
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You're saying that you think ##2 < \frac 3 2##?
No, but I am quite confused--we are talking about x, not ##\left| x-1 \right|##, right?
 
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  • #10
PeroK
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No, but I am quite confused--we are talking about x, not ##\left| x-1 \right|##, right?
Yes, sorry, of course it's inside the radius of convergence. In any case, what does ##x = 2## tell you about ##A_n##?
 
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  • #11
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You're saying that you think ##2 < \frac 3 2##?
Can you please tell me what I should do next? This is not a homework question and I am doing it just for interest.
 
  • #12
PeroK
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Can you please tell me what I should do next? This is not a homework question and I am doing it just for interest.
I've actually told you: put ##x = 2##.
 
  • #13
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Yes, sorry, of course it's inside the radius of convergence. In any case, what does ##x = 2## tell you about ##A_n##?
When ##x \geq 2 \: or \: x \leq 0##, the series converges.
 
  • #14
PeroK
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When ##x>2 \: or \: x<0##, the series converges.
And what is the series when ##x = 2##?
 
  • #15
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And what is the series when ##x = 2##?
Converges at x=2?
 
  • #16
PeroK
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Converges at x=2?
Write down the sequence for ##x = 2##.
 
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  • #17
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Write down the sequence for ##x = 2##.
$$\sum_{n=0}^ \infty{A_n}$$
 
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  • #18
PeroK
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$$\sum_{n=0}^ \infty{A_n}$$
That's what you should have done after post #2.
 
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  • #19
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That's what you should have done after post #2.
Oh, thank you! So it is actually convergent because the limit is less than 1 if we sub in x=2!
 
  • #20
PeroK
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Oh, thank you! So it is actually convergent because the limit is less than 1 if we sub in x=2!
In general, the series:
$$\sum_{n = 0}^{\infty} A_n $$
converges, if and only if the power series:
$$\sum_{n = 0}^{\infty} A_n x^n$$
converges at ##x = 1##.

Those two properties are clearly equivalent.
 
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