$\cosh(z) := \sum_{k=0}^\infty \frac{z^{2k}}{(2k)!}, z \in \mathbb{C}$ and $\cosh(z) = \frac{1}{2} (\exp(z)+\exp(-z))$

Question

Let $\cosh$ be given by

$$\cosh(z) := \sum_{k=0}^\infty \frac{z^{2k}}{(2k)!}, z \in \mathbb{C}$$

How can one show that the power series converges absolutely for all $z \in \mathbb{C}$?

I also have to show that

$$\cosh(z) = \frac{1}{2} (\exp(z)+\exp(-z))$$

When using the ratio test I get

$$\left|\frac{(2k)!}{(2k+1)!} \frac{z^{2k+1}}{z^{2k}}\right| = \frac{|z|}{k+1} \Rightarrow 0 \text{ (n} \to \infty) < 1 \text{ for all }z \Rightarrow r = \infty$$ (while $r$ is the convergence radius). So, the power series converges absolutely for all $z$. Is this correct?

Regarding $$\cosh(z) = \frac{1}{2} (\exp(z)+\exp(-z))$$ I have that,

\begin{eqnarray} \frac{1}{2}\left(\sum_{k=0}^{+\infty}\frac{z^k}{k!} + \sum_{k=0}^{+\infty}\frac{(-1)^kz^k}{k!}\right) &=& \frac{1}{2}\sum_{k=0}^{+\infty}\frac{(1 -(-1)^k)z^k}{k!} \\ &=& \frac{1}{2}\sum_{k=0}^{+\infty}\frac{(1 -(-1)^{2k})z^{2k}}{(2k)!} + \frac{1}{2}\sum_{k=0}^{+\infty}\frac{(1 -(-1)^{2k})z^{2k}}{(2k+1)!} \\ &=& 0 + \frac{2}{2}\sum_{k=0}^{+\infty}\frac{z^{2k}}{(2k+1)!}\\ &=& \sum_{k=0}^{+\infty}\frac{z^{2k}}{(2k+1)!} \end{eqnarray}

Is that correct?

Answer 1

In the first step you changed $1+(-1)^{k}$ to $1-(-1)^{k}$ and that is why you are getting a wrong answer.