Prove $\sum_0^\infty x^k/k!$ does not converge uniformly to $e^x$ on the entire real line R.

Question

Prove $\sum_0^\infty x^k/k!$ does not converge uniformly to $e^x$ on the entire real line R.

I know that this power series converges absolutely and uniformly over any compact interval, [$a,b$], however, I am unsure how to prove that this does not hold beyond this interval.

Answer 1

Hint: if $f_n \to f$ uniformly on $\mathbb R$, then $f - f_n$ must be bounded for some $n$.

Answer 2

By Taylor series with Lagrange remainder $$ e^x=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{(e^x)^{(2n)}|_t}{2n!}x^{2n}=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{e^t}{2n!}x^{2n} $$ where $|t|<|x|$. If it converges uniformly to $e^x$on the entire real line, then given $\epsilon>0$, there is $N$ such that for any $n>N$ and $x\in\Bbb{R}$, there is $$ \left|e^x-\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}\right|=\left|\frac{e^t}{2n!}x^{2n}\right|<\epsilon $$ But this is not true for if fix $n$, we can always find $x$ large enough ($e^{-t}\geqslant1$) so that $$ \left|x^{2n}\right|>\frac{2n!}{e^t}\epsilon\quad\text{and }\quad\left|\frac{e^t}{2n!}x^{2n}\right|>\epsilon $$ So this series doesn't converge uniformly to $e^x$on the entire real line $\Bbb{R}$.