How to show that $\sum_{k=0}^{\infty} \frac{x^{k}}{k!}$ represents a continuous function

Question

(This is a homework problem) I am trying to show that the series $\sum_{k=0}^{\infty} \frac{x^{k}}{k!}$ represents a continuous function on $\mathbb{R}$. My idea was to show that the functions $f_{k}(x) = \frac{x^{k}}{k!}$ are continuous and that the series converges uniformly. I was able to show that the $f_{k}(x)$ are continuous, but I'm not sure about the second part. My main concern is that the $f_{k}(x)$ are not uniformly convergent for $k \geq 2$, so it seems that the series will not be uniformly convergent (I haven't shown this formally yet, however).

Answer 1

Pick any $B>0$. Then, if $|x|\leq B$, then one has that $|f_k(x)|=|x|^k/k!\leq B^k/k!$ for each non-negative integer $k$. Since $$\sum_{k=0}^{\infty}\frac{B^k}{k!}=\exp(B)$$ is convergent, Weierstrass's $M$-test reveals that the series $\sum_{k=0}^{\infty} x^k/k!$ converges uniformly for $x\in[-B,B]$. Together with the continuity of the partial sums $x\mapsto\sum_{k=0}^K f_k(x)$ for each non-negative integer $K$, this implies that the limit function is continuous on $[-B,B]$. Since $B$ can be made arbitrarily large, it follows that the limit function—which is, of course, $x\mapsto\exp(x)$—is continuous on the whole line.

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Remark: One can't conclude that the convergence of the power series is uniform on the whole line—only that it is uniform on each compact subinterval of it. However, this suffices to prove continuity on the whole line. Indeed, a function is continuous if and only if it is continuous at each point. Then, for any point $x\in\mathbb R$, one can take $B>0$ so large that $x\in[-B,B]$ and use the previous argument to conclude that the limit function is continuous at $x$.

Answer 2

We will use the fact that $$ \sum_{n=0}^\infty\frac{x^n}{n!} $$ converges absolutely for all $x\in\mathbb{R}$. This can easily be shown using the ratio test and means that for any $x$, there is an $N$ so that $$ \sum_{n=N+1}^\infty\frac{|x|^n}{n!}\le\epsilon $$ Find an $N$ so that we have $$ \sum_{n=N+1}^\infty\frac{|x|^n}{n!}\le\epsilon\qquad\text{and}\qquad\sum_{n=N+1}^\infty\frac{|x+h|^n}{n!}\le\epsilon $$ Using the Mean Value Theorem, for some $\xi$ between $x$ and $x+h$, $$ \begin{align} |f(x+h)-f(x)| &\le\left|\,\sum_{n=0}^N\frac{(x+h)^n-x^n}{n!}\,\right| +\sum_{n=N+1}^\infty\frac{|x|^n}{n!} +\sum_{n=N+1}^\infty\frac{|x+h|^n}{n!}\\ &\le h\left|\,\sum_{n=0}^N\frac{\xi^{n-1}}{(n-1)!}\,\right|+2\epsilon\\ \end{align} $$ Thus, choosing $0\lt h\le1$ so that $$ h\max_{|\xi-x|\le 1}\left|\,\sum_{n=0}^N\frac{\xi^{n-1}}{(n-1)!}\,\right|\le\epsilon $$ we have that $$ |f(x+h)-f(x)|\le3\epsilon $$

Answer 3

A very simple proof of this actually involves complex analysis. The set of complex numbers extends to the reals.

$e^{z}=\sum_{0}^ \infty \frac{z^{k}}{k!} $

This is a series for $e^{z}$ expanded at $\alpha=0$.

In complex analysis in order for a number to be inside the disc of convergence it has to follow the following conditions:

Let $\beta$ be the closest singularity.

$|z-\alpha|<|\alpha-\beta |$

Since $e^z$ contains no singularities all numbers are a set of the disc of convergence therefore $e^z$ is entire which is not difficult to figure out. This means it is analytic at all points, which means it is differentiable at all points which means it is continous at all points in the complex plane.