# How to show that $\sum_{n=1}^{\infty}\frac{x^n}{n!}$ does not converge uniformly on the set $(0,+\infty)$?

Show that the series does not converge uniformly: $$\sum_{n=1}^{\infty}\frac{x^n}{n!},\ \ \ x\in E=(0,+\infty)$$

Well, here is what I did: $$\frac{x^n}{n!}=u_n(x)\Rightarrow \lim_{n\rightarrow\infty}\frac{u_{n+1}(x)}{u_n(x)}=\lim_{n\rightarrow\infty}\frac{x}{n+1}=0\ \ \ \forall x\in E \Rightarrow\\ \Rightarrow\sum_{n=1}^{\infty}\frac{x^n}{n!}\ \ \text{converges by the ratio test.}\\ \text{Now let's check uniform convergence:}\\ r_n(x)=S(x)-S_n(x)=\sum_{k=n+1}^{\infty}\frac{x^k}{k!},\ \ \lim_{n\rightarrow\infty}\sup_{x\in E}|r_n(x)|\ne0\Rightarrow\\ \Rightarrow \sum_{n=1}^{\infty}\frac{x^n}{n!}\ \ \text{does not converge uniformly.}$$ The question is how to prove that $$\lim_{n\rightarrow\infty}\sup|r_n(x)|\ne0$$?

• For fixed $x$ we have $\limsup_{n\to\infty}|r_n(x)| = 0$. Oct 10 '19 at 15:35
• May I ask what this is for? you're trying to prove it doesn't converge when the ratio test pretty solidly shows otherwise Oct 10 '19 at 15:36
• @AlexRobinson Emphasis is on "uniformly on $(0,\infty)$". Oct 10 '19 at 15:36
• @amsmath Note that the OP is computing $\lim_{n\rightarrow\infty}\sup_{x\in(0,\infty)}$ and not $\limsup_{n\rightarrow\infty}$. Oct 10 '19 at 15:43
• @MichaelBurr I see. Oct 10 '19 at 15:45

With $$f_m(x)=\sum_{n=0}^mx^n/n!$$ and $$f(x)=\lim_{m\to \infty}f_m(x)$$ we have for $$m>0$$ that $$\sup_x|f_m(x)-f(x)|\ge |f_m(m)-f(m)|=\sum_{n=m+1}^{\infty}m^n/n!>$$ $$>m^{m+1}/(m+1)!=(m/(m+1))\cdot(m^m/m!)\ge (m/(m+1)\ge 1/2.$$

Observe that $$r_n(x)\geq\frac{x^{n+1}}{n!}.$$ If the series were to converge uniformly, for any $$\varepsilon>0$$, there would be some $$N\in\mathbb{N}$$ such that $$|r_n(x)|<\varepsilon$$ for all $$n>N$$ and $$x$$. Since $$\lim_{x\rightarrow\infty}r_n(x)\geq\lim_{x\rightarrow\infty}\frac{x^{n+1}}{n!}=\infty,$$ this does not happen and the convergence is not uniform.

Note, alternatively (following the setup in the question), $$\lim_{n\rightarrow\infty}\sup_{x\in(0,\infty)}|r_n(x)|\geq\lim_{n\rightarrow\infty}\sup_{x\in(0,\infty)}\frac{x^{n+1}}{n!}=\lim_{n\rightarrow\infty}\infty.$$

• But where does the first inequality come from? Oct 10 '19 at 15:41
• @Bonrey Everything is positive and a sum of positive numbers is at least as large as any one in the sum. Oct 10 '19 at 15:42

If a series of functions $$\sum u_n(x)$$ converges uniformly, then its finite sums $$S_N(x) = \sum_{i=1}^N u_n(x)$$ are uniformly Cauchy. In particular $$\Vert S_{n+1} - S_n \Vert_\infty$$ converges to zero. Which is not the case here as $$\frac{x^n}{n!}$$ is unbounded on $$[0, \infty)$$.

And if you want to go precisely in the direction of your question, you have:

$$r_n(2n+2) \ge \frac{(2n+2)^{n+1}}{(n+1)!} \ge \frac{2n+2}{n+1} \ge 2$$