Show $\sum\limits_{n=0}^\infty (-1)^n\dfrac{z^{2n+1}}{(2n+1)!}$, represent holomorphic function on $\mathbb{C}$.
(Weierstrass test) Let $f_n : I \to \mathbb{R}$ be a sequence of functions with $|f_n(x)| \le M_n$ for all $x \in I$ and $k=1,2, \dots$. If $\sum_n M_n$ converges then $\sum_n f_n$ converges uniformly on $I$.
I get the idea is to find $M_n$ such that $$\left|(-1)^n\dfrac{z^{2n+1}}{(2n+1)!}\right| = \dfrac{1}{(2n+1)!}|z|^{2n+1}\leq |z|^{2n+1}$$ is less than. Do I assume that $|z|\leq R$? If so, the only way the geometric series formula works is if I restrict $R$ even more. Is that correct?
Pretty sure that's the series for sine. This doesn't actually converge uniformly on the whole complex plane I don't think. But, you can do something a little tricky. You take just a disk of some radius. It will converge uniformly ON THAT DISK, and so you can prove it is analytic on that disk as the limit of a uniformly converging sequence of analytic functions. Since being analytic is a local condition, this prove that the function is analytic at any point in the complex plane (not including infinity).