How do you prove the series $\sum_\limits{n=0}^{\infty}(2n+1)(-x)^n$ is convergent? The Wikipedia article about Formal Power Series states that
$$
S(x)=\sum_{n=0}^{\infty}(-1)^n(2n+1)x^n,
$$
if considered as a normal power series, has radius of convergence 1. How do we prove this?
I understand that for positive $x$ the series is alternating, and apart from a finite number of terms the sequence $(2n+1)x^n$ is monotone decreasing in $n$ for $0\le x<1$ and tends to zero for $n\to\infty$. Thus the alternating series test applies and the series is (conditionally) convergent for $0\le x<1$. But, since this is a power series, it has to be convergent in a disk in the complex plane, and thus we conclude the series is in fact convergent for complex $x$ when $|x|<1$. Is this argument correct?
Also, I checked with Maple and I get:
$$
S(x)=\frac{\sqrt{\pi}(-x)^{1/4}\text{LegendreP}\left(\frac12,\frac12,\frac{1-x}{1+x}\right)}{(1+x)^{3/2}},
$$
where the function in the numerator is the Legendre function of the first kind. This closed form solution, if correct, suggests that there is a branch cut for $x>0$ and $x<-1$, hence I would naively think based on this representation that true converge occurs for $-1<x<0$. Does anybody know how to prove this identity and where on the complex plane it holds?

The root test for me,
since I know that
$\lim_{n \to \infty} a^{1/n}
=\lim_{n \to \infty} n^{1/n}
=1$
for any $a > 0$.
Then
$\lim_{n \to \infty} (2n+1)^{1/n}
\le \lim_{n \to \infty} (3n)^{1/n}
\le \lim_{n \to \infty} 3^{1/n}\lim_{n \to \infty} n^{1/n}
=1
$
and
$\lim_{n \to \infty} (2n+1)^{1/n}
\ge 1$
so the radius of convergence is
$1$.
