Not so long ago, I've tried to learn the concept of uniform convergence and I must admit, my understanding is quite unpolished and based only on epsilon-delta definition.
About the problem itself, this is what I have to investigate:
$$\sum_{n=0}^\infty \frac{(-1)^n(x^{2n+1})}{(2n+1)!}$$
And I did it already with [-1,1] interval:
$$\frac{|(-1)^n(x^{2n+1})|}{|(2n+1)!|} = \frac{(x^{2n+1})}{(2n+1)!}$$
$x^{2n+1}$ is at most 1 on [-1,1], so:
$$\frac{(x^{2n+1})}{(2n+1)!} \le \frac{1}{(2n+1)!}$$
Which is then found convergent by ratio test:
$$\frac{1 * (2n+1)!}{(2n+3)! * 1} = \frac{1}{(2n+2)(2n+3)}$$
So, I'm actually confused with what I should do to investigate the uniform convergence on the interval (0, +$\infty$), as the same trick obviously won't do there. I've also noticed that $\sum_{n=0}^\infty \frac{(-1)^n(x^{2n+1})}{(2n+1)!} = \sin x$ by Tailor series, and I am sure that I've missed something really simple, yet important.
If it converges uniformly on $(0,\infty)$ then the general term would tend to $0$ uniformly. So there exist $N$ such that $|\frac {(-1)^{n}x^{2n+1)}} {(2n+1)!}| <1$ for all $n >N$ and for all $x$. You get a contradiction to this by letting $x \to \infty$.