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

(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).

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.

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 $x$.

• Is there a way to prove convergence on the whole line explicitly? Or is this method of proving convergence on compact subintervals and extending the length of the intervals to $\infty$ the "standard" way to deal with this? – Roflcakzorz May 19 '15 at 7:07
• @Roflcakzorz I'm afraid that convergence is not uniform on the whole line, that's why you can't use the argument based on uniform convergence directly. The argument I propose provides a convenient workaround: restrict attention to compact subintervals, where uniform convergence does obtain, and conclude that the limit function is continuous on each compact interval. This, in turn, implies, that it is continuous on the whole of $\mathbb R$. – triple_sec May 19 '15 at 7:12
• @Roflcakzorz Upon reflection, yes, I believe that the general method of (i) restricting attention to “smaller” sets, (ii) proving the desired result for such sets, and (iii) then trying to extend the result to “larger” sets is a “standard” practice that turns up under different guises throughout real analysis (and also related fields like measure theory and general point-set topology). – triple_sec May 19 '15 at 7:16
• I see, thank you very much for your detailed explanation :) – Roflcakzorz May 19 '15 at 7:58

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 $$|f(x+h)-f(x)|\le3\epsilon$$

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.