# Absolute Convergence question

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#3

(Original post by

I think I've done this wrong, but I don't know how to start it.

**Airess3**)I think I've done this wrong, but I don't know how to start it.

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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#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

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.

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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(Original post by

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.

**0x2a**)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.

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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#7

**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.

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#9

(Original post by

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.

**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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(Original post by

Is that what you were having trouble with, or was it some other part of the process?

**davros**)Is that what you were having trouble with, or was it some other part of the process?

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#11

(Original post by

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".

**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".

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#13

(Original post by

Yeah, sorry that's what I meant.

**Airess3**)Yeah, sorry that's what I meant.

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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(Original post by

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

**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

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