I am trying to wrap to figure out how to better understand how to more rigorously compute this. My attempt was to use:
Theorem: For $\Sigma\ a_nx^n$. Let $\beta=\limsup |a_n|^{1/n}$ and $R=1/\beta$. Then we have convergence for $|x|<R$.
Taking a fairly basic example, consider $\Sigma_{n=0}^{\infty}\ \frac{(-1)^n x^{2n+1}}{2n+1}$. Then we have that $\beta=\limsup |a_n|^{1/n}=\limsup |\frac{1}{(2n+1)^{1/n}}|=\limsup(\frac{1}{3},\frac{1}{\sqrt5},\frac{1}{\sqrt[3]7},...)=\frac{1}{3}$. Thus the series is convergent for $|x|<3$. However, I believe this is the series expansion for $\arctan(x)$ which is supposed to converge for $|x|\le1$. I'm not really sure where I went wrong. Also, if that is correct, how would I then go about showing the behavior of the series at the endpoints of $R$? Do you just set $x$ equal to the value and check to see if the new series converges/diverges?
Here we have $|a_{2n}| = 0$ and $|a_{2n+1}| = 1/(2n+1)$. Thus,
$$\limsup_{n \to \infty}|a_n|^{1/n} = \limsup_{n \to \infty}|a_{2n+1}|^{1/(2n+1)} = \exp\left(\lim_{n \to \infty} \frac{-\log(2n+1)}{2n+1}\right) = \exp(0) = 1$$