Is the Taylor series uniformly convergent on $\Bbb R$?

Question

How can we Prove that $\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ does not converge uniformly on $\mathbb{R}$? By using weierstrass-M test, it is easy to show that this series converges uniformly on a compact interval. Is it true to say that:\ By contradiction, suppose $\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ converges uniformly to $e^x$, then $(s_{n+1}-s_n)$ converges uniformly to zero, however, $\dfrac{x^{n+1}}{(n+1)!}$ is not bounded on $R$?

Answer 1

Hint: If $\sum f_n$ converges uniformly on a set $E,$ then $\sup_E|f_n|\to 0.$

Answer 2

Suppose that the series converges uniformly on $\Bbb R$. In particular, taking $\epsilon = 1$, $\exists n_0$ s.t. for $n\ge n_0$, $x\in\Bbb R$: $$\left|\sum_{k=0}^{n}\frac{x^k}{k!} - e^x\right|\le 1.$$ But this is impossible because at least two reasons:

(1) When $x\to+\infty$ the exponential growths faster than any polynomial.

(2) When $x\to-\infty$ the exponential $\to 0$ while the polynomial...