Showing the power series of $\cos(x)$ converges uniformly to $\cos(x)$ on every bounded interval

Show that the power series of $\cos(x)$ converges uniformly to $\cos(x)$ on every bounded interval.

My attempt: The power series for $\cos(x)$ is $$\sum_{n=0}^{\infty} \frac{{(-1)^{n}}{x^{2n}}}{(2n)!}$$ A sequence of functions $f_n: X \rightarrow Y$ converges uniformly if there exists a function $f: X \rightarrow Y$ such that for every $\epsilon > 0$ there is an $N$ such that $d_Y(f_n(x),f(x)) < \epsilon$ for every $x \in X$ and every $n \geq N$. In our problem, $$f_n = \sum_{j=0}^{n} \frac{{(-1)^{j}}{x^{2j}}}{(2j)!}, \qquad f = \cos(x).$$ Using the ratio test I got the radius of convergence is equal to infinity. Can I then conclude that the power series converges uniformly to $\cos(x)$ on every bounded interval because the radius of convergence is $(-\infty,\infty)$?

• That depends on what you know about power series. If you know a power series converges compactly on its (open) disk of convergence, you're done. If not, you need to prove it at least in the special case of $\cos$. – Daniel Fischer Mar 16 '14 at 15:00

• For every $x$ such that $|x|\leqslant K$ and every $N\geqslant1$, $$\left|\sum_{n\geqslant N} \frac{(-1)^nx^{2n}}{(2n)!}\right|\leqslant\sum_{n\geqslant N}\frac{K^{2n}}{(2n)!}\leqslant\frac{K^{2N}}{(2N)!}\sum_{i\geqslant0} \frac{K^{2i}}{(2N)^{2i}}.$$
• For every $N\geqslant2K$, $$\frac{K^{2N}}{(2N)!}\leqslant\left(\frac{\mathrm e}4\right)^{2N},\qquad\sum_{i\geqslant0} \frac{K^{2i}}{(2N)^{2i}}\leqslant2.$$