Uniform convergence of the exponential series on a bounded interval 
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. 
 A: 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.
A: Suppose that $|x| \leq M$, then for any $N$,
$|\sum_{n=1}^N \frac{x^n}{n!}| \leq \sum_{n=1}^N \frac{M^n}{n!} \to e^M$.  So the series converges uniformly by the Weierstrass M-test.
A: As suggested, the same question tackled with Weierstrass:
Proof
We will use the Weierstrass M-Test  as observed in Kosmala $8.4.11$. We will have to compare our sequence $f_n=\frac{x^n}{n!}$ to some other sequence $M_n$. We have a bounded interval, so let us use this fact and pick  the bound $B$ such that $|x| \leq B$ for all $x$ in the interval,  this leads to the comparison:
          $$ |f_n(x)|=\left| \frac{x^n}{n!} \right| = \frac{|x^n|}{|n!|}=\frac{|x|^n}{n!}  \leq \frac{B^n}{n!}$$
         These numbers are certainly nonnegative, we now need to verify that $\sum M_n$ converges, where $M_n=\frac{B^n}{n!} $, luckily in chapter $7$ we discussed that this representation is actually the exponential function, evaluated at the value $B$. We thus realise that:
         $$ \sum_{n=0}^\infty\frac{B^n}{n!} = \exp(B)= e^B.  $$
         This means that the series converges and we know exactly to which value. By the Weierstrass $M$-test we now know that   $\sum _ {k=0} ^\infty
   \frac{x^
    k}
   {k!} $ converges uniformly and absolutely on the interval. $\square$
