Show that $\sum_{k=0}^\infty \frac{(−1)^k}{(2k+1)!} x^{2k+1}$ converges uniformly on $[−a,a]$ for every $a>0$ , but not uniformly on R.

Question

Show that $\sum_{k=0}^\infty \frac{(−1)^k}{(2k+1)!} x^{2k+1}$ converges uniformly on $[−a,a]$ for every $a>0$ , but not uniformly on R.

I know that the Weierstrass-M Test must be used and that $$\sum_{k=0}^\infty \frac{(−1)^k}{(2k+1)!} x^{2k+1} = \sin(x)$$ but I'm not sure where to go from there.

Answer 1

Weierstrass $M$-test can be used in the following way: $$\left|\frac{(−1)^k}{(2k+1)!} x^{2k+1}\right|=\frac{|x|^{2k+1}}{(2k+1)!},$$ if $x$ lies in an interval of the form $[-a,a]$, then we can bound this by $\frac{a^{2k+1}}{(2k+1)!}$, and the series having this as a general term converges.

For the uniform convergence on $\mathbb R$, we can use the following general fact: if a series of functions $\sum_{n=1}^{+\infty}f_n$ converges uniformly on $\mathbb R$, then the sequence $\left(f_n\right)_{n\geqslant 1}$ converges uniformly to $0$ on $\mathbb R$.