Show that the uniform convergence of $f_n(x)=x-\frac{x^n}{n}$ on $0 \leq x \leq 1$?

Question

Show that the following sequence of $f_n(x)$ converges uniformly on the interval $[0,1]$

My work:

$$f_n(x)=x-\frac{x^n}{n}$$

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

$$\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}x-\frac{x^n}{n}$$

The sequence converges for $|x|<1$

$$\lim_{n\rightarrow \infty} x-\frac{x^n}{n}=x-0$$

$$f(x)=x$$

$$|x-\frac{x^n}{n}-x|< \epsilon$$

$$\frac{x^n}{n}< \epsilon$$

Does this question need Lambert function?

I am not sure how to proceed... Any help will be appreciated.

Answer 1

Given $\epsilon > 0$, choose a positive integer $N > \frac{1}{\epsilon}$. For all $n > N$ and $x\in [0,1]$, $$\left\lvert\left(x - \frac{x^n}{n}\right) - x\right\rvert = \frac{x^n}{n} \le \frac{1}{n} < \frac{1}{N} < \epsilon$$

Thus $f_n(x)$ converges uniformly to $x$ on $[0,1]$.

Answer 2

Since $\lim_{n\to\infty}\frac{x^n}n=0$, if your sequence converges uniformly to some function, then that function must be $f(x)=x$. But$$f_n(x)-f(x)=-\frac{x^n}n$$and $(\forall x\in[0,1]):f_n(x)-f(x)\in\left[-\frac1n,0\right]$. Therefore, $(\forall x\in[0,1]):\left|f_n(x)-f(x)\right|\leqslant\frac1n$ and, since $\lim_{n\to\infty}\frac1n=0$, the sequence $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$.