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 $\lim_{n\rightarrow\infty}\sup|r_n(x)|\ne0$?
Best Answer
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 $$\sup_x|f_m(x)-f(x)|\ge |f_m(m)-f(m)|=\sum_{n=m+1}^{\infty}m^n/n!>$$ $$>m^{m+1}/(m+1)!=(m/(m+1))\cdot(m^m/m!)\ge (m/(m+1)\ge 1/2.$$