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?