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?
 A: First of all, note that your closed form seems a little off. A much simpler closed form is
$$S(x) = \frac{1-x}{(x+1)^2}$$
However, it is trivial to find the convergence in the complex plane even without this. We apply the Ratio Test, which is stronger than the alternating series test in that it can show absolute convergence. Note that the ratio test says that, for a series $\sum_{k=0}^\infty a_n$, the series coverges absolutely if $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$. Working this out, we find that
$$\lim_{n\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{n\to\infty}\left|\frac{(-1)^{n+1}(2n+3)x^{n+1}}{(-1)^n(2n+1)x^n}\right|$$
$$=\left|\lim_{n\to\infty}\frac{(2n+3)x^{n+1}}{(2n+1)x^n}\right|=|x|$$ 
Thus, we have that the series converges when $|x|<1$, as you stated in your argument. To answer your question, your argument is more or less correct, and this confirms is. The boundary $|x|=1$ is an entirely different ballgame, especially working on the complex plane, and must be checked separately, which I leave to the OP if desired.
A: By ratio test
$$\lim_{n\to+\infty}\frac{|(-1)^{n+1}(2n+3)x^{n+1}|}{|(-1)^n(2n+1)x^n|}=|x|$$
thus if $|x|<1$, the series converges.
and
if $|x|>1$, it diverges.
the convergence radius is $R=1$.
A: The Ratio Test is the simplest way to show that the radius of convergence is $1$.  For $|x|=1$, the terms don't go to $0$, so there is no convergence on the boundary of the disk.
A: 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
$1$.
