Proving Series of Functions on (-1,1)

[Whistlekins]

Homework Statement

Prove that the series \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1} is well-defined and differentiable on (-1,1).

Homework Equations

The Attempt at a Solution

I know that the function is the series expansion of arctan(x), but that it not we are showing here (however it asks in a later question to show that it is). I don't know what it means by "well-defined", but I'm going to guess it means continuous and convergent on its domain. I am guessing that I should use the Uniform Cauchy Criterion to show that it converges uniformly, and thus showing that it is differentiable.

But I'm not sure how to show that for all ε > 0, there exists and N ≥ 1 such that, for all n, p ≥ N and all x in (-1,1), |f_n(x) - f_p(x)| < ε

Also, how would I show that it is actually the series expansion for arctan(x)?

A nudge in the right direction would be greatly appreciated.

 

[clamtrox]

I am guessing that I should use the Uniform Cauchy Criterion to show that it converges uniformly, and thus showing that it is differentiable.

But I'm not sure how to show that for all ε > 0, there exists and N ≥ 1 such that, for all n, p ≥ N and all x in (-1,1), |f_n(x) - f_p(x)| < ε

That sounds like a good plan. Can you see when |f_n(x) - f_p(x)| would be maximized? You can bound this from above by choosing a value for x that gives the largest difference.

Also, how would I show that it is actually the series expansion for arctan(x)?

You can probably show that their derivatives are equal.

 

[Whistlekins]

That sounds like a good plan. Can you see when |f_n(x) - f_p(x)| would be maximized? You can bound this from above by choosing a value for x that gives the largest difference.

Can you expand on what you mean by maximized? Would that happen when n = 0 and p -> ∞?

 

[clamtrox]

Can you expand on what you mean by maximized? Would that happen when n = 0 and p -> ∞?

I mean choose value for n, then pick a p that maximizes the difference for arbitrary x and then choose x to find the absolute upper bound for a given n. And notice that the series is alternating, so n=0 and p→∞ is not the maximum.