Uniform converges of $f_n(x)=x-\frac{x^n}{n!} , x\in [-a,a] ,a>0$

Question

I have 2 questions :

1.with the given following function series :

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

I need to show that for $x\in [-a,a]$ $ ,a>0 $ there is $P,Q>0$ such that :

$|f_n(x)-f_{n-1}(x)|\leq PQ^{n-1}\frac{|x|^n}{n!}$

and I stack at this point :

$$\begin{align} |f_n(x)-f_{n-1}(x)|&=\left|x-\frac{x^{n+1}}{(n+1)!}-\left(x-\frac{x^{n}}{n!}\right)\right|\\&=\left|x-\frac{x^{n+1}}{{n+1}!}-x+\frac{x^{n}}{n!}\right|\\&=\left|\frac{x^n}{n!}-\frac{x^{n+1}}{(n+1)!}\right|\\&=\left|\frac{x^n}{n!}\left(1-\frac{x}{n+1}\right)\right|\\&=\left|\frac{x^n}{n!}\right|\left|1-\frac{x}{n+1}\right|\\&=\frac{|x|^n}{n!}\left|1-\frac{x}{n+1}\right|\end{align} $$

so I need to prove that :

$|1-\frac{x}{n+1}|\leq PQ^{n-1}$

then I tried : $|1-\frac{x}{n+1}|\leq 1+|\frac{x}{n+1}|\leq 1+\frac{|x|}{n+1} \leq 1+\frac{a}{n+1} $

from here I stack.

My second question is to show that $f_n(x)$ converges unifomly at $x\in [-a,a]$ $ ,a>0 $

Answer 1

Fixed $x\in [-a, a]$

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

Since $[\lim_{n\to \infty }\frac{ x^n}{n!}=0]$

Hence, $f_n\to f$ pointwise on $[-a, a]$ where $f:[-a, a]\to \Bbb{R}$ defined by $f(x) =x$

Claim : $f_n\to f$ uniformly on $[-a, a]$ i.e $(f_n) \to f$ in the space $(C[-a, a], \|•\|_{\infty}) $

$\begin{align}\|f_n-f\|_{\infty}&=sup\{|f_n(x)-f(x)|:x\in [-a, a]\}\\&=sup\{|x-\frac{x^n}{n!}-x| : x\in[-a,a]\}\\&=sup\{\frac{|x|^n}{n!}:x\in[-a,a]\}\\&=\frac{a^n}{n!} \to 0 [\text{ as $n\to\infty$}]\end{align}$

Hence, $f_n\to f$ uniformly on $[-a, a]$