Csdrhrt

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.

Let $A$ be the bounded interval and suppose that $|x|leq B$ for $xin A$. Note that for $xin A$
$$
leftlvertsum_{k=m+1}^nfrac{x^k}{k!} rightrvertleqsum_{k=m+1}^n leftlvertfrac{x^k}{k!}rightrvertleqsum_{k=m+1}^nfrac{B^k}{k!}
$$
so
$$
sup_{xin A}leftlvertsum_{k=m+1}^nfrac{x^k}{k!} rightrvertleqsum_{k=m+1}^nfrac{B^k}{k!}to 0tag{1}
$$
as $m, nto infty$ since the last series converges. It follows that the partial sums of $sum_{n=0}^inftyfrac{x^n}{n!}$ are uniformly Cauchy.

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.

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:
$$ sum_{n=0}^inftyfrac{B^n}{n!} = exp(B)= e^B. $$
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. $square$