Show: The function series
$$\sum _ {k=0} ^\infty
\frac{x^
k}
{k!} $$
converges uniformly on each bounded interval in $\mathbb{R}$.
Discussion I think a good approach will be to deploy the Cauchy Criterion for uniform convergence. Our definition of the Cauchy Criterion from class ( variation on Kosmala theorem 8.4.6.) is as follows:
Let $ \{ f_n \}$ be sequence of functions defined on $D$, if $$\forall \varepsilon >0, \exists n_0, \text{whenever } n,m \geq n_0 \qquad ||f_n -f_m ||_\infty < \varepsilon $$ (where $n>m$),
then $\sum f_n$ is uniformly convergent.
Suppose we denote the function sequence of the sum by $f_n$, so for some arbitrary $n$ and $m$ we have that $n>m$ and we consider:
$$|f_n(x) -f_m(x) |=\Bigg|\sum _ {k=0} ^n
\frac{x^
k}
{k!} -\sum _ {k=0} ^m
\frac{x^
k}
{k!} \Bigg|= \Bigg|\sum _ {k=m+1} ^n
\frac{x^
k}
{k!} \Bigg|$$
Now notice that we are dealing with a bounded interval, so there must exist some upper bound $B$, that is larger than any element $x$ in this interval. We can use this bound to estimate:
$$\Bigg|\sum _ {k=m+1} ^n
\frac{x^
k}
{k!} \Bigg| \leq \Bigg|\sum _ {k=m+1} ^n
\frac{B^
k}
{k!} \Bigg|$$
But I don't quite know how to finish the proof $\dots$, basically I want to be able to make this as small as possible ($\varepsilon$), because then we will have shown uniform convergence, we then of course take the supremum in the end.
Best Answer
Let $A$ be the bounded interval and suppose that $|x|\leq B$ for $x\in A$. Note that for $x\in A$ $$ \left\lvert\sum_{k=m+1}^n\frac{x^k}{k!} \right\rvert\leq\sum_{k=m+1}^n \left\lvert\frac{x^k}{k!}\right\rvert\leq\sum_{k=m+1}^n\frac{B^k}{k!} $$ so $$ \sup_{x\in A}\left\lvert\sum_{k=m+1}^n\frac{x^k}{k!} \right\rvert\leq\sum_{k=m+1}^n\frac{B^k}{k!}\to 0\tag{1} $$ as $m, n\to \infty$ since the last series converges. It follows that the partial sums of $\sum_{n=0}^\infty\frac{x^n}{n!}$ are uniformly Cauchy.