# $f_n(x)=\sum_{k=0}^n \frac{x^{2k}}{(2k)!}$ converges uniformly on any compact subset of $\Bbb R$

Consider a sequence of functions on $\Bbb R$ defined by $$f_n(x)=\sum_{k=0}^n \frac{x^{2k}}{(2k)!}, n \ge 0.$$ I'd like to show that $\{f_n(x)\}$ converges uniformly on any compact subset of $\Bbb R$. We have $\{f_n(x)\}$ is a monotonically increasing sequence which converges pointwise to a continuous function $f(x)=\frac{e^x+e^{-x}}2$ since $$\lim_{n \to \infty}f_n(x)=\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}=\frac{e^x+e^{-x}}2,$$ by Dini's theorem $\{f_n(x)\}$ converges uniformly.

How does the proof look? Is there any better proof?

• Weierstrass $M$-test can be used. – Winther Oct 3 '16 at 14:57
• That's a fine proof. – Thomas Andrews Oct 3 '16 at 15:04
• Do you mean we prove $\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}$ converges uniformly by Weierstrass M-test, so does its partial $f_n(x)$? – user Oct 3 '16 at 15:05
• "$\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}$ converges uniformly" means that the partials converge uniformly. @user – Thomas Andrews Oct 3 '16 at 15:22
• $f_n(x)$ is the partial sum of $\sum_{k=0}^n \frac{x^{2k}}{(2k)!}$. To prove $\{f_n(x)\}$ converges uniformly, we can prove that $\sum_{k=0}^n \frac{x^{2k}}{(2k)!}$ converges uniformly by using Weierstrass M-test, can't we? – user Oct 3 '16 at 15:26

That's a fine proof.

A more direct proof is to show it is uniformly convergent on $[-N,N]$ for any $N$.

Given $\epsilon>0$, pick $n_0$ so that $\frac{N^{2n_0}}{(2n_0)!}<\frac{\epsilon}{2}$ and $\frac{N^2}{(2n_0+1)(2n_0+2)}<\frac{1}{2}$. (You need to argue why you can find such $n_0$.)

Then for $n\geq n_0$ and $x\in[-N,N]$ you show that:

$$\left|\frac{e^x+e^{-x}}{2}-f_n(x)\right|\leq \frac{\epsilon}{2}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)=\epsilon$$

The Weierstass $M$-test proof is to define $a_k(x)=\frac{x^{2k}}{(2k)!}$. Then, for $x\in [-N,N]$, $|a_k(x)|\leq a_k(N)=M_k$ so $$\sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}$$ converges uniformly on $[-N,N]$ because $\sum_{k=0}^\infty M_k$ converges to $\frac{e^{N}+e^{-N}}{2}$.

• Very instructive! +1 – Mark Viola Oct 3 '16 at 17:39