Pointwise and uniform convergence of $\sum_{n=0}^\infty \frac{x^n}{e^{nx}}, n \in \mathbb{N}$ $$\sum_{n=0}^\infty \frac{x^n}{e^{nx}}, n \in \mathbb{N}, D=[0, +\infty) \rightarrow \mathbb{R}$$
I tried using Weierstrass M-test
$\frac{x^n}{e^{nx}} =({\frac{x}{e^{x}}})^n \le (\frac{1}{e})^n$ and $|\frac{1}{e}|\lt1$, so I thought because $(\frac{1}{e})^n$ converges (geometric series), that
$\sum_{n=0}^\infty \frac{x^n}{e^{nx}}$ converges absolutely and uniformly and therefore pointwise.
Is that correct and if not, where are my mystakes?
 A: The denominator in $x/e^x$ is nowhere $0,$ so this function is defined for every $x\in\mathbb R.$
$$
\frac d{dx} (xe^{-x}) = e^{-x}(1 - x) \begin{cases} >0 & \text{if } x<1, \\ <0 & \text{if } x>1. \end{cases}
$$
Thus this function increases on $(-\infty,1]$ and decreases on $[1,\infty),$ and thus has an absolute maximum at $x=1,$ where its value is $1/e < 1.$
Thus the conclusion of the argument showing $\left|\dfrac x {e^x}\right| \le \dfrac 1 e < 1$ is correct when $x>0.$
However $\dfrac x {e^x} \to -\infty$ as $x\to-\infty.$ When $x=0$ then the inequality $\left|\dfrac x {e^x}\right| \le \dfrac 1 e < 1$ clearly holds, and thus when $x$ is near $0$ it holds, but just how far you can go in the negative direction and still have that inequality hold is a question that will bear further examination.
Let $x_0$ be the one real number for which $x_0 e^{-x_0}=-1.$ The series diverges for $x=x_0$ and for $x<x_0.$ And it cannot converge uniformly on $[x_0,\infty),$ for reasons that may become clear when you see why the assertions above hold. But try the Weierstrass M-test on the interval $[x_1,\infty)$ where $x_1$ is any number at all that is $>x_0.$
