How can I prove $\sum_{n=1}^{\infty} \frac{x^n}{n!}$ converges uniformly on $[-a,a], a>0$ but not on $\mathbb{R}$? I'm looking to show that:

$\sum_{n=1}^{\infty} \frac{x^n}{n!}$ converges uniformly on $[-a,a]$ for $a>0$, but does not converge uniformly on $\mathbb{R} $. 

How can I do this? My book mentions it as part of an example but doesn't elaborate on it and I can't intuitively see it at the moment. Thanks for any help!
 A: I assume that the second part of the question is what you're struggling with. To prove that a sequence $\{f_n\}$ does not converge uniformly to $f$, you want to show that

For some $\epsilon > 0$, and for all $N \in \Bbb N$: there exists an $n \geq N$ and an $x$ (in $\Bbb R$) such that $|f_n(x) - f(x)|\geq \epsilon$ 

For any $n$, try to show that a suitable $x$ exists.
A: On $[-a,a]$, $|x^n/n!| \le a^n/n!$ for all $n\in \Bbb N$, and $\sum_{n = 1}^\infty a^n/n!$ converges. By the Weierstrass $M$-test, the series $\sum_{n = 1}^\infty x^n/n!$ converges uniformly on $[-a,a]$. 
To see that $\sum_{n = 1}^\infty x^n/n!$ does not converge uniformly on $\Bbb R$, note that for every positive integer $N$, 
$$\left|\sum_{n = N}^{2N} \frac{(2N)^n}{n!}\right| >  \frac{(2N)^N}{(2N)!}(N+1) \ge 2.$$
Therefore, by the Cauchy criterion for uniform convergence, the series $\sum_{n = 1}^\infty x^n/n!$ does not converge uniformly on $\Bbb R$.
A: That a sequence of functions $f_n(x)$ converges pointwise to $f(x)$ means that for any $\epsilon > 0$ and any $x$ you can find an $N_x$ such that for any $n > N$, we have $|f_n(x) - f(x)| < \epsilon$.
For uniform convergence, on the other hand, you have to find one such $N$ that works for all values of $x$. This is possible in the bounded case, but if we look at the unbounded domain $\Bbb R$, then the further out on the real line you go, the further out in the sequence you have to go to get $f_n$ close to $f$. For any given candidate for $N$, if you use large enough values for $x$, you won't have $|f_{N+1}(x) - f(x)| < \epsilon$ any more, and this is what prohibits this sequence from being uniformly convergent.
