# Absolute Convergence question

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#1
I think I've done this wrong, but I don't know how to start it.
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5 years ago
#2
Are you trying to find the radius of convergence?
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5 years ago
#3
(Original post by Airess3)
I think I've done this wrong, but I don't know how to start it.
This looks like a radius of convergence question. It is usually best to solve these questions using the ratio test (possibly with a bit of tweaking/common sense, but this is not necessary here).

I'll also note that you have some very messed up manipulation of exponents in your working. I know this is basic AS material, but if you can't do it well and reliably you're setting yourself up for failure with questions like this.
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5 years ago
#4
I'm not sure what you mean by finding the convergence, but I have spotted a working mistake if that helps. When you took out the 3 from the denominator, the (-1) power should be on the 3 taken out, not (x^n/3^2n)
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5 years ago
#5
If you're trying to find the radius of convergence, then use the ratio test/root test as suggested above.

Say you're using the root test, then to find the radius of convergence, you want to just look at the sequence itself where is your power series.

So perform the root test as such: . Then your radius of convergence should be and .

Or with the ratio test, let , then your radius of convergence should then be again.
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#6

For the first paper working, the third last line, I meant to write 1^2 instead of 1^-1.

Tried both, but have no idea how to do the second one. The professor said the answer is R=9 because of 1/3 x 1/3. And IxI is greater than or equal to 9. And it converges absolutely. My answers are nowhere near that.
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5 years ago
#7
(Original post by Airess3)

For the first paper working, the third last line, I meant to write 1^2 instead of 1^-1.

Tried both, but have no idea how to do the second one. The professor said the answer is R=9 because of 1/3 x 1/3. And IxI is greater than or equal to 9. And it converges absolutely. My answers are nowhere near that. 0
#8
Could I ask you how you did that? I thought if you did subtracted the powers away from each other then you'd get 1^(-2). And if you substitute infinity into the n's then the answer would be 1.
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5 years ago
#9
(Original post by Airess3)
Could I ask you how you did that? I thought if you did subtracted the powers away from each other then you'd get 1^(-2). And if you substitute infinity into the n's then the answer would be 1. Is that what you were having trouble with, or was it some other part of the process?
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#10
(Original post by davros) Is that what you were having trouble with, or was it some other part of the process?
Yes, I understand it now, thanks! The question also asked about the range of values of real numbers x for which the series converges. The answer for that is IxI is greater than or equal to 9. But I don't know why it's "greater than or equal to".
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5 years ago
#11
(Original post by Airess3)
Yes, I understand it now, thanks! The question also asked about the range of values of real numbers x for which the series converges. The answer for that is IxI is greater than or equal to 9. But I don't know why it's "greater than or equal to".
Are you sure it's not ??
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#12
Yeah, sorry that's what I meant.
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5 years ago
#13
(Original post by Airess3)
Yeah, sorry that's what I meant.
So are you happy with the question now?

Note that your series is basically just a multiple of a GP with common ratio x/9, so you'd expect the answer to be |x/9| < 1 or |x| < 9
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#14
(Original post by davros)
So are you happy with the question now?

Note that your series is basically just a multiple of a GP with common ratio x/9, so you'd expect the answer to be |x/9| < 1 or |x| < 9
Yeah, I understand it now.
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