Prove the series $\sum u_n(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots$ converges uniformly on every bounded subset of $\Bbb R$.

Question

Let $[-t,t]$ be the bounded subset of $\mathbb R$ for $t>0$.

Let $\varepsilon$ be given. Then $\exists$ $ m\in \mathbb N$ such that for any $p \in \mathbb N$ & $x \in[-t,t]$, we get

$$\left| \frac{x^n}{n!}\right| < \left| \frac{t^n}{n!}\right|<\frac \varepsilon p \quad \forall n\ge m. \tag 1$$

Now considering,

$$ \begin{align} \left| u_{n+p}-u_n\right| & =\left| \frac{x^{n+1}}{(n+1)!}+\frac{x^{n+2}}{(n+2)!}+\frac{x^{n+3}}{(n+3)!}+\cdots+\frac{x^{n+p}}{(n+p)!} \right| \\[8pt] & {} \le \left| \frac{x^{n+1}}{(n+1)!}\right| + \left| \frac{x^{n+2}}{(n+2)!}\right| + \left| \frac{x^{n+3}}{(n+3)!}\right| +\cdots+ \left| \frac{x^{n+p}}{(n+p)!} \right| \tag 2 \end{align} $$

(where $u_n=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+ \cdots +\frac{x^n}{n!}$)

From (1) & (2) $\forall n\ge m$ & $\forall p\in \mathbb N$, we have

$$\left| \frac{x^{n+1}}{(n+1)!}\right|+ \left| \frac{x^{n+2}}{(n+2)!}\right| + \left| \frac{x^{n+3}}{(n+3)!}\right| +\cdots+ \left| \frac{x^{n+p}}{(n+p)!} \right| < \frac{\varepsilon}{p} \cdot p= \varepsilon$$

Hence by Cauchy's Principle $\sum u_n(x)$ converges uniformly on $[-t,t]$. Since $t>0$ is arbitrary $\sum u_n(x)$ converges uniformly on any bounded subset of $\mathbb R$.

Is My Solution Correct? Can this be done by Weierstrass' M-test? Why this series is not uniformly convergent on $\mathbb R$?

Answer 1

A power series $\sum_n a_n z^n$ with radius of convergence $R$ converges uniformly on $\{z: |z| \le r\}$ if $r < R$. In this case $R = \infty$.

Answer 2

Your solution is not correct. You state an inequality for every $p$ (after having chosen m) and that is wrong. You may use Weierstrass M-test. Given your $t>0$ choose an integer $m > 2t$. Then $$ \sum_{n\geq m} \frac{t^n}{n!} \leq \frac{t^m}{m!} \sum_{k\geq 0} \left( \frac{t}{m}\right)^k \leq\frac{t^m}{m!} \sum_{k\geq 0} 1/2^k \leq 2\frac{t^m}{m!} $$

Answer 3

Hint: Use the $M$-test and then the ratio test.

The function converges uniformly on every bounded subset of $\mathbb{R}$. In Particular, fix a radius of convergence, R.

Then $$|\frac{x^n}{n!}| \leq \frac{R^n}{n!}$$

Let $A_n=\frac{R^n}{n!}$, and now you need only show thaat $\sum_{n \geq 0} A_n$ converges.

Try the ratio test, which asserts that if $\lim_{n \to \infty}\frac{A_{n+1}}{A_n}=L<1$, then the series will converge absolutely, morally since it will eventually be less than a convergent geometric series.