For what x does $P_c(x) = \sum^\infty_{n=0} (-1)^{n+1}\cdot n\cdot x^n$ converge?

Question

For what x does the exponential series $P_c(x) = \sum^\infty_{n=0} (-1)^{n+1}\cdot n\cdot x^n$ converge?

What I got so far:

$\sum^\infty_{n=0} (-1)^{n+1}\cdot n\cdot x^n = (-1)\sum^\infty_{n=0} (-1)^{n}\cdot (\sqrt[n]n)^n\cdot x^n = (-1)\sum^\infty_{n=0} ((-1)\cdot (\sqrt[n]n)\cdot x)^n$

We know that $|((-1)\cdot (\sqrt[n]n)\cdot x)|$ must be < 1 in order to converge (geometric sum):

$ |((-1)\cdot (\sqrt[n]n)\cdot x)| < 1$

Case 1: $ ((-1)\cdot (\sqrt[n]n)\cdot x) < 1 \Leftrightarrow (-1)\cdot (\sqrt[n]n)\cdot x < 1 \Leftrightarrow (-1)\cdot x < \frac{1}{(\sqrt[n]n)} \Leftrightarrow x > \frac{-1}{(\sqrt[n]n)}$

Case 2: $ ((\sqrt[n]n)\cdot x) < 1 \Leftrightarrow (\sqrt[n]n)\cdot x < 1 \Leftrightarrow x < \frac{1}{(\sqrt[n]n)}$

We know that $x < |\frac{-1}{(\sqrt[n]n)}| = |\frac{1}{(\sqrt[n]n)}| = \frac{1}{(\sqrt[n]n)} \le \pm 1 $

So $x < (-1) < 1$. Is this way right, must $x$ be $< -1$ in order for $P_c(x) $ to converge? Is there a simpler way to get to a solution? Thanks in advance!

Answer 1

Cauchy-Hadamard:

$$\lim_{n\to\infty}\sqrt[n]{\left|(-1)^{n+1}n\right|}=\lim_{n\to\infty}\sqrt[n]n=1$$

so the series converges for $\;|x|<1\;$ .

Answer 2

May be, you could consider that $$\sum^\infty_{n=0} (-1)^{n+1} n x^n=x\sum^\infty_{n=0} (-1)^{n+1} n x^{n-1}=x \frac{d}{dx} \left(\sum^\infty_{n=0} (-1)^{n+1} x^n \right)=-x\frac{d}{dx} \left(\sum^\infty_{n=0} (-x)^{n} \right)$$ Now, in brackets, you have a simple geometric sum easy to compute (and analyze, for sure).