Let $f_n: [0,1]\rightarrow \mathbb{R}$ be a sequence of functions defined by $$f_n(x)=\frac{x^n}{n!}.$$ Prove that $f_n(x)$ converges uniformly to $0$.
We know that $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ is this enought to prove that $f_n \rightarrow 0$?
On
The answer by lorenzo was what I had in mind with my comment. But if you really want to beat this problem with a hammer, then here goes: \begin{align} \sum_{n=0}^{\infty}\sup_{x\in[0,1]}\left|\frac{x^n}{n!}\right| &=\sum_{n=0}^{\infty}\frac{1}{n!} = e < \infty. \end{align} Hence, by the vanishing criterion for convergent series, it follows that $\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}\left|\frac{x^n}{n!}\right| = 0$, which is precisely the definition (or atleast it's very easy to see this is equivalent) of uniform convergence to $0$ over the the interval $[0,1]$.
Let $\varepsilon>0$: then $|f_n(x) - 0|=|\frac{x^n} {n!} |\leq \frac{1}{n!}<\varepsilon$ for $n>N:=\frac{1}{\varepsilon}$. Since $\varepsilon>0$ is arbitrary we have shown that $\forall\varepsilon>0$ there exists $N$ such that $|f_n(x) - 0|<\varepsilon\ \forall n>N$ thus $(f_n) $ converges uniformly to $0$ (on $[0,1]$) by definition of uniform convergence.