How Does Abel's Theorem Apply to Uniform Convergence?

[sandra1]

Hi everyone,

I'm reading my analysis textbook and trying to prove Abel's Theorem but I don't really get it. Any help would be very much appreciated.

Thank you very much!

Homework Statement

Suppose that R in (0, ∞) is the radius convergence for the power series ∑(k=0 to inf) a_k(x-a)^k.

a. if ∑(k=0 to ∞) (a_k)*R^k converges, then ∑(k=0 to ∞) a_k(x-a)^k converges uniformly on [a-R+ ε1, a+R] for any ε1>0.
b. if ∑(k=0 to ∞) (a_k)* (-R)^k converges, then ∑(k=0 to ∞) a_k(x-a)^k converges uniformly on [a-R, a+R-ε2] for any ε2 >0.

Homework Equations

The Attempt at a Solution

I know this has to do with the Weierstrass M-Test but I'm confused about the [a-R+ε1, a+R] and [a-R+ε2, a+R] parts:

So since R is radius of convergence , for |x-a|<R the series ∑(k=0 to ∞) a_k(x-a)^k converges, and so:
|a_k(x-a)^k| < (a_k)* (R)^k
and ∑(k=0 to ∞) (a_k)*R^k converges, so by the Weierstrass M-test ∑(k=0 to ∞) a_k(x-a)^k converges uniformly.

Could you tell me what do I need to do with connect with the [a-R+ε1, a+R] and [a-R+ε2, a+R]?

 

[Office_Shredder]

You need the epsilon because the series might converge for a+R, but diverge for a-R. Hence the M-test actually fails on the interval [a-R,a+R] when comparing the series to the value at a+R

 

[sandra1]

You need the epsilon because the series might converge for a+R, but diverge for a-R. Hence the M-test actually fails on the interval [a-R,a+R] when comparing the series to the value at a+R

Thank you. So here's my trying. Would you mind looking through if there's any problem with it? I also have problem with the -R in part b.

a. For R in (0, ∞), choose x in [a-R+ε1, a+R] with an ε1>0 so (x-a) <=R. Therefore:
|(a_k)(x-a)^k| <= |(a_k)(R^k)|
But since ∑(k=0 to ∞) (a_k)*R^k converges as given, by Weierstrass M-Test for uniform convergence => ∑(k=0 to ∞) a_k(x-a)^k converges uniformly on [a-R+ε1, a+R]

b. For R in (0, ∞), choose x in [a-R,a+R-ε2] with an ε1>0. With|x-a| <=R:
|(a_k)(x-a)^k| <= |(a_k)((R)^k)|
We know that (a_k)* (-R)^k converges only converges iff |(a_k)* (-R)^k| converges, which means (a_k)* (R)^k converges, so by Weierstrass theorem we also have ∑(k=0 to ∞) a_k(x-a)^k converges uniformly on [a-R, a+R-ε2].