Question: is the sequence of function $(f_n)$ defined by, $f_n(x)=\sum_{k=0}^n \frac{x^k}{k!}$ for all $x\in A$ where $A$ is bounded subset of $\mathbb{R}$, is uniformly convergent on $A$?
My attempt: consider
$|f_{n+1}(x)-f_n(x)|=|\frac{x^{n+1}}{(n+1)!}|= \frac{1}{(n+1)!}|x^{n+1}|$
$$<\epsilon$$ if $(n+1)!>\frac{1}{\epsilon} |x^{n+1}|$
i.e. if $(n+1)n!>\frac{1}{\epsilon} |x^{n+1}|$
i.e. if $n+1>\frac{1}{\epsilon (n!)} |x^{n+1}|$
i.e if $n>\frac{1}{\epsilon (n!)} |x^{n+1}|-1$
So let $N\text{ be an integer} ≥\frac{1}{\epsilon (n!)} |x^{n+1}|-1$ then
$|f_{n+1}(x)-f_n(x)|<\epsilon$ for all $n≥N$
Hence $(f_n)$ converges uniformly on $A$.
Is am i correct?
Further, I am confused why such an integer $N$ exists? is because of $A$ is bounded subset of $\mathbb{R}$ hence $|x^{n+1}|$ is bounded for every $x\in A$ and hence such an integer exists! But still I have doubt on existence of $N$.
Please help...
Let $A$ be a bounded subset of $\mathbb{R}$. Put $M=\sup_{x\in A}|x|$. Then for every $n$, if $x\in A$ then $\frac{|x^n|}{n!}\leq \frac{M^n}{n!}$. Since the numerical series $\sum_{n=1}^{\infty}\frac{M^n}{n!}$ converges, it follows from Weierstrass M test that the series $\sum_{n=1}^{\infty}\frac{x^n}{n!}$ converges uniformly on the set $A$.