Let $\{a_n\}$ is sequence of bounded numbers and $$f_n(x)=a_0+a_1x+\frac{a_2x^2}{2!}+...+\frac{a_nx^n}{n!}.$$ Then $\{f_n\}$ is uniformly convergent on $[-\alpha, \alpha]$ for and $\alpha>0.$
First I tried to find the limit function, but I failed.
\begin{align*} \lim f_n(x)&=\lim (a_0+a_1x+\frac{a_2x^2}{2!}+...+\frac{a_nx^n}{n!}) \\ & < M\lim (1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}) = Me^x.\end{align*} But how can I find the limit function, or is there any way to show uniform convergence without finding limit function?
One can prove that the sequence $\left(f_n\right)_{n\geqslant 1}$ is uniformly Cauchy: let $m> n$ and $x\in [-\alpha,\alpha]$. Then $$ \lvert f_m(x)-f_n(x)\rvert=\left\lvert \sum_{j=n+1}^ma_jx^j/j!\right\rvert $$ and using the triangle inequality, the fact that $\lvert a_jx^j\rvert\leqslant \sup_{k\geqslant 1}\lvert a_k\rvert\cdot \alpha^j$ and the series $\sum_{j\geqslant 1}\alpha_j/j!$ converges gives that $$ \lim_{m,n\to +\infty}\sup_{x\in [-\alpha,\alpha]}\lvert f_m(x)-f_n(x)\rvert=0. $$