Show that $f_n(x)=\frac{1}{x+n}$ converges uniformly on $0\leq x \leq 1$?

Question

Consider the following sequence $\left\{f_n\right\}$ defined on $0 \leq x \leq 1$. Show that the sequence converges uniformly on $0 \leq x \leq 1$

$$f_n(x)=\frac{1}{x+n}(n=1,2,3,...) $$

My work:

First notice that the sequence is continuous for every values of $n$.

$$f(x)=\lim_{n \rightarrow \infty}f_n(x)$$

$$\lim_{n \rightarrow \infty}f_n(x)=\lim_{n \rightarrow \infty}\frac{1}{x+n}$$

$$\lim_{n \rightarrow \infty}f_n(x)=0$$

If the function converges uniformly on the interval $[0,1]$ then there exists an $\epsilon>0$ for ever $n>N$

$$|f_n(x)-f(x)|< \epsilon$$

$$|\frac{1}{x+n}-0|<\epsilon$$

$$\frac{1}{x+n}<\epsilon$$

$$n>\frac{1}{\epsilon}-x$$

Hence we can say that

$$N=\frac{1}{\epsilon}-x$$

Can someone help me out. It is my first time doing something like this.

Answer 1

We have that

$$x\in [0,1]\implies \dfrac{1}{x+n}\le \dfrac 1n.$$ Thus, for any $\epsilon>0$ there exists $N\in\mathbb{N}$ ($N>1/\epsilon$) such that

$$n\ge N\implies |f_n(x)-f(x)|= \dfrac{1}{x+n}\le \dfrac 1n\le \dfrac1N<\epsilon.$$

Answer 2

Uniform convergence is the statement that the sequence $s_n \to 0$ as $n\to \infty$, where

$$ s_n = \operatorname{sup}_{x \in [0, 1]} | f_n(x) - f(x) | $$

Since $f_n$ is monotonic decreasing on $[0, 1]$, it achieves its maximum at $x = 0$, and so $s_n \leq f_n(0) = \frac{1}{n}$. So indeed, the limit of the sequence $s_n$ is $0$.