Calculating Improper Integral w/ Power Series of r=1

aaaa202
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At exam today I was to calculate an improper integral of a function f defined by a power series.
The power series had radius of convergence r=1.
Inside this radius you could of course integrate each term, i.e. symbologically:
∫Ʃ = Ʃ∫
The only problem is that the improper integral went from 0 to 1.
Is it then true that:
limx->1[∫Ʃ ]=limx->1[Ʃ∫]
and what theorem assures this? At the exam I didn't think about this unfortunately, but I would probably not have known what to do anyways. I think there is a theorem called Abels theorem which shows that a power series is continuous also in x=±r, but I'm not sure if that's what I am looking for.
 
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Are you familiar with uniform convergence?

Abel's theorem basically says that if a series converges at x = R, it converges uniformly over the interval
0 ≤ x ≤ R.

As a consequence, you can also observe uniform convergence over the interval -R ≤ x ≤ 0.

A corollary would be that it converges on -R ≤ x ≤ R.
 
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Yes I am familiar with uniform convergence. And you agree that from what I told you I would have had to invoke Abels theorem to give a satisfactory answer?
 
aaaa202 said:
Yes I am familiar with uniform convergence. And you agree that from what I told you I would have had to invoke Abels theorem to give a satisfactory answer?

Uniform convergence would be a sufficient condition for you to be able to switch limits/derivatives/integrals around. So yes, Abel's theorem could be used here as it guarantees the uniform convergence over the whole interval [-R,R].
 
I don't think you need anything fancy at all. Assuming that your variable x is "integrate from 0 to x" then for every x<1, the thing you have written in square brackets [ ] is equal whether the integration or the summation comes first, because you're integrating over [0,x] which is bounded away from 1. You just did algebra inside the limit which is legal for the values of x that you are considering
 
Im switching 2 limits:
limx->1limn->∞[Ʃanxn] = limn->∞limx->1[Ʃanxn]
I only know that the power series converges for lxl<1
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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