What is the radius convergence of the series? 
Find the radius convergence of $\sum_{n = 0}^{\infty} \frac{x^{2n+1}}{2n+1}$

Let's start with:
$$f(x) = \sum_{n = 0}^{\infty} \frac{x^{2n+1}}{2n+1} = \int \sum_{n = 0}^{\infty} x^{2n} \,dx = \int \frac{1}{1-x^2}\,dx$$
Now, evaluating the integral, we get:
$$f(x) = \frac{1}{2}\, \ln \left| \frac{1+x}{1-x} \right| + C$$
Now, I know that the radius convergence of $\sum_{n=1}^\infty x^{2n}$ is $R=1$, because it's a geometric series, we demand: $\left| x^2 \right| < 1 \implies \left| x \right| < 1$.
BUT, we actually want to know when the integral converges, aren't we?
So I have a closed form of $f(x)$ and I was able to notice that at $x=\pm 1$ $f(x)=\pm \infty$.
I think we should check when $\frac{1+x}{1-x} > 0$. This happens if and only if:
$$(1+x > 0 \, \text{and} \, 1-x > 0) \, \text{or} \, (1+x < 0 \, \text{and} \, 1-x < 0)$$
The second argument is always false, so we have that $\frac{1+x}{1-x} > 0$ iff $-1 < x < 1$.
Hence, $R=1$.
I'd be glad to get a review of my work, and maybe getting an alternative approach if mine is wrong or tedious (Was I able to figure out what $R$ is in an early phase?)
Thanks.

The series converges trivially if $x = 0$.
If $x \neq 0$, we can apply the Ratio Test.
\begin{align*}
\lim_{n \rightarrow \infty} \frac{\dfrac{x^{2n + 3}}{2n + 3}}{\dfrac{x^{2n + 1}}{2n + 1}} & = \lim_{n \rightarrow \infty} \frac{x^{2n + 3}}{2n + 3} \cdot \frac{2n + 1}{x^{2n + 1}}\\
             & = \lim_{n \rightarrow \infty} x^2 \cdot \frac{2n + 1}{2n + 3}\\
             & = x^2 \lim_{n \rightarrow \infty} \frac{2 + \frac{1}{n}}{2 + \frac{3}{n}}\\ 
             & = x^2\\
\end{align*}
which is less than $1$ when $|x| < 1 \Rightarrow -1 < x < 1$.  
If $x = 1$, then we obtain
\begin{align*}
\sum_{n = 0}^{\infty} \frac{1}{2n + 1} & = 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} + \cdots\\
& > 1 + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10} + \frac{1}{12} + \frac{1}{14} + \frac{1}{16} + \cdots\\
& > 1 + \frac{1}{4} + \color{blue}{\frac{1}{8}} + \color{blue}{\frac{1}{8}} + \color{green}{\frac{1}{16}} + \color{green}{\frac{1}{16}} + \color{green}{\frac{1}{16}} + \color{green}{\frac{1}{16}} + \cdots\\
& = 1 + \frac{1}{4} + \color{blue}{\frac{1}{4}} + \color{green}{\frac{1}{4}} + \cdots\\
& = \infty
\end{align*}
so the series diverges.
If $x = -1$, then 
\begin{align*}
\sum_{n = 0}^{\infty} \frac{(-1)^{2k + 1}}{2k + 1} & = -1 - \frac{1}{3} - \frac{1}{5} - \frac{1}{7} - \frac{1}{9} - \frac{1}{11} - \frac{1}{13} - \frac{1}{15} - \cdots\\
& = -1 - \left(\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} + \cdots\right)\\
& < -1 - \left(\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \cdots\right)\\
& = -\infty
\end{align*}
so the series diverges.
Thus, the series converges if and only if $-1 < x < 1$. 
