# 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?

• why $x/e^x < 1/e$? – Yanko Feb 9 '18 at 14:51
• This is related math.stackexchange.com/questions/405087/… – Yanko Feb 9 '18 at 14:52
• You forgot to specify the domain of $x$. Is $x$ supposed to be positive? – Winther Feb 9 '18 at 15:01
• This is a geometric series with ratio $\frac{x}{e^x}$. I think this should converge if and only if $|\frac{x}{e^x}| < 1$. – Anu Feb 9 '18 at 15:05
• @Anu : That establishes only pointwise convergence, i.e. convergence for each value of $x$ separately, among those for which $|x/e^x| < 1.$ It fails to say whether the convergence is uniform. – Michael Hardy Feb 9 '18 at 15:08

## 1 Answer

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 $>x_0.$

• Sorry about forgetting the domain, I added it in the first post. Would be my way of using the M-test be correct then to show uniform convergence? – lolan496 Feb 9 '18 at 16:50
• @lolan496 : Yes. You have uniform convergence on every interval of the form $[x_1,\infty)$ where $x_1>-1,$ but you don't have uniform convergence on $[-1,\infty). \qquad$ – Michael Hardy Feb 9 '18 at 18:07