# 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.

• Just use the weierstrass m-test. It’s basically Cauchy criterion but all the work is already done for you. – Shalop Dec 17 '18 at 22:05
• oh wow, you're right. – Wesley Strik Dec 17 '18 at 23:15
• Weierstrass just kills this problem right off... – Wesley Strik Dec 17 '18 at 23:35

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.

• Why does the sum go to zero? – Wesley Strik Dec 17 '18 at 22:55
• Since $\sum_{n=0}^\infty \frac{B^n}{n!}$ converges (to $\exp(B)$) it follows that the partial sums of this series are Cauchy and hence the rightmost sum of $(1)$ goes to zero as $m, n\to \infty$. – Foobaz John Dec 17 '18 at 23:03
• Why not use Weierstrass M? – zhw. Dec 17 '18 at 23:14
• My question was how I could fix this proof, Foobaz answered my question. – Wesley Strik Dec 17 '18 at 23:36

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