How to prove a series is convergent in Complex Plane

Question

Prove that $\sum_{n=0}^{\infty}(-1)^{n}z^{2n}/({2n})! $ is convergent for all $z ∈ \mathbb{C}$

I am familiar how to tackle this if it were a Real Analysis problem, but unsure how using Leibniz's test changes the result given that it's Complex.

I see how this is $cos(z)$ but how I don't see how to explicitly prove that it is convergent using conventional methods.

Answer 1

The ratio and root test apply as normal. Indeed, from these tests we define the radius of convergence of a power series $\sum a_nx^n$ as

$$\rho:=\frac1{\limsup_{n\to\infty}\sqrt[n]{|a_n|}}$$

and when the limit exists

$$\rho=\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}$$

This definition is directly derived from these tests when you clear the $x$. The radius mean that the series converges if $|x|<\rho$. Sometimes it converges too when $|x|=\rho$. Take a look at this.

Answer 2

Well let's have ($a_n$) a complex sequence verifying: $|\frac{a_{n+1}}{a_n}| \rightarrow 0$ , with $a_n \neq0 $ ,for n integer

Then for a sufficient N, we have: $n>N \implies |\frac{a_{n+1}}{a_n}|<\frac{1}{2}$

Hence you can prove that : $|\frac{a_{n}}{a_N}| < \frac{1}{2^{n-N}} $ , for any n>N

Which means ($\sum a_n$) converges. Of course this stays true for a series of functions, given that the ratio condition is true for every z.

So you can use that