Showing $\dfrac{e^{-nx}}{n} < \epsilon$

Question

I am trying to prove that $f_n(x)=\dfrac{e^{-nx}}{n}\rightarrow f(x)=0$ uniformly on $x\in[0, \infty)$. To do this I am trying to show that: $$\dfrac{e^{-nx}}{n} < \epsilon$$ I tried to do it this way, but I am not sure if it is correct: \begin{align} &\dfrac{e^{-nx}}{n} \\ &<\dfrac{1}{n} \quad \forall n>N_1 \ \exists N_1\in \mathbb{N} \\ &<\epsilon \quad \forall n>N_2 \ \exists N_2\in \mathbb{N} \\ \\ \therefore \dfrac{e^{-nx}}{n}&<\epsilon \quad \forall n>N=\max\{N_1, N_2\} \end{align}

This in turn implies uniform convergence. Is this the correct way to do it?

Answer 1

Let $n \geq 1$ then $g(x)=e^{-nx}$ is a decreasing function defiend on $[0,+\infty)$

Thus $e^{-nx} \leq e^0=1$ thus $$\sup_{x \geq 0}|\frac{e^{-nx}}{n}| \leq \frac{1}{n}$$

Thus is $\epsilon >0$ then for $n \geq n_0=[\frac{1}{\epsilon}]+1$ we have that $$ \sup_{x \geq 0}|\frac{e^{-nx}}{n}| \leq \epsilon$$

Answer 2

You got the idea.

Need to show that for $\epsilon >0$ there is a

$N \in \mathbb{Z^+}$ such that for $x \in [0,\infty)$ and

$n \gt N$ we have $|\dfrac{e^{-nx}}{n} -0| \lt \epsilon.$

$\dfrac{e^{-nx}}{n} =\dfrac{1}{ne^{nx}} =$

$\dfrac{1}{n(1+nx+(nx)^2/2!+...)} \le \dfrac{1}{n}.$

Archimedes :

There is a $N \in \mathbb{Z^+}$ such that

$N >1/\epsilon.$

For $x \in [0,\infty )$ and $n >N$ we have

$\dfrac{e^{-nx}}{n} \le 1/n \lt 1/N \lt \epsilon.$

Answer 3

It is enough to say

$$\frac{e^{-nx}}n\le\frac1n<\epsilon$$ for all $n>\dfrac1\epsilon$.

---

In fact, to prove uniform convergence, it suffices to say

$$\frac{e^{-nx}}n\le\frac1n\text{ and }\dfrac1n\text{ converges}.$$