Prove that $\sum_{n=0}^{\infty}\frac{x^n}{n!}$ is not uniformly convergent on $(0, +\infty)\ni x$.

Question

Prove that $\sum_{n=0}^{\infty}\frac{x^n}{n!}$ is not uniformly convergent on $(0, +\infty)\ni x$.

I wanted to do it by applying Cauchy's test and coming to a contradiction (i.e. $\epsilon>$ positive constant):

$$\epsilon>|\frac{x^{n+1}}{(n+1)!}+\frac{x^{n+2}}{(n+2)!}+\cdots+\frac{x^{n+p}}{(n+p)!}|$$

$$\epsilon>|\frac{x^{n+1}}{(n+1)!}\cdot(1+\frac{x}{n+2}+\cdots+\frac{x^{p-1}}{(n+2)\cdots(n+p)})|.$$

The expressions within the ||'s reminded me of $e^x=1+x+\cdots$, but after several tries I still wasn't able to evaluate. I've also tried standard method of 'find the smallest one in the brackets and take it $p$ times', but that didn't lead anywhere.

Another idea where I took $x$ to be an expression involving $n$ worked, but is it even legal? $x$ is supposed to be fixed, and $n$ is arbitrarily large, so I don't know.

Hints or full answers, I will be very grateful.

Answer 1

Note that

$$\sup_{x \in (0,\infty)}\left|\sum_{k=n+1}^\infty \frac{x^k}{k!}\right| \geqslant\sup_{x \in (1,\infty)}\sum_{k=n+1}^{2n} \frac{x^k}{k!}\geqslant \sup_{x \in (1,\infty)}\frac{nx^n}{(2n)!} $$

Can you show that the RHS does not converge to $0$ as $n \to \infty$?

Answer 2

Much easier if we can use that the sum is the exponential: the partial sum is a polynomial and the exponential grows faster than any polinomial, so for any $n$ $$\sup_{x\in(0,\infty)}\left|e^x - \sum_{k=0}^n\frac{x^k}{k!}\right| = \infty$$