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.

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

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  • $\begingroup$ Why does the sum go to zero? $\endgroup$ – Algebra geek Dec 17 '18 at 22:55
  • $\begingroup$ 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 $m, n\to \infty$. $\endgroup$ – Sri-Amirthan Theivendran Dec 17 '18 at 23:03
  • $\begingroup$ Why not use Weierstrass M? $\endgroup$ – zhw. Dec 17 '18 at 23:14
  • $\begingroup$ My question was how I could fix this proof, Foobaz answered my question. $\endgroup$ – Algebra geek 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.

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As suggested, the same question tackled with Weierstrass:


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. $\square$

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