# Evaluating power series on the radius of convergence for $\tan^{-1}(x)$

Suppose that I know the series for $(\tan^{-1})'=\sum_{n=0}^{\infty} (-1)^nx^{2n}$ converges when $\lvert x\rvert \lt 1$.

Then I know that $$\tan^{-1}(x)=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{2n+1}$$ only for $\lvert x \rvert \lt 1$.

My professor asks me:

How do I know that $$\lim_{x\to\ 1}\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \lim_{x\to\ 1}\frac{(-1)^nx^{2n+1}}{2n+1}\;?$$

In general, once the radius of convergence has been found for a power series, I think I can rest assured the series obtained by differentiating or integrating term by term has the same radius of convergence. What more do I need to consider to talk about the points on the radius?

• It's a lot more subtle, because the symbols don't interchange if you replace $\lim_{x \to 1}$ with $\lim_{x \to -1}$. – user296602 Apr 16 '18 at 20:28
• Funnily enough, my professor also describes this as "subtle". Where should I begin reading? – Jungleshrimp Apr 16 '18 at 20:32
• Have you heard about Abel, but not this one killed by Cain? – Przemysław Scherwentke Apr 16 '18 at 20:32
• I have heard "of" Abel – Jungleshrimp Apr 16 '18 at 20:34
• en.wikipedia.org/wiki/Abel%27s_theorem – Chappers Apr 16 '18 at 20:35