Algebra Involving Factorials and Complex Numbers

Question

Let $n \in \mathbb{Z}^{\ge 0}$ and $z \in \mathbb{C}$. Why is the following true? $$\frac{(iz)^{2n}}{(2n)!} + \frac{(iz)^{2n+1}}{(2n+1)!} = \frac{(iz)^{n}}{n!}$$ I’m following a proof of Euler’s formula and comprehend everything but this last step. The furthest I’ve gotten is: $$\frac{(iz)^{2n}}{(2n)!} + \frac{(iz)^{2n+1}}{(2n+1)!} = \frac{(iz)^{2n}((2n + 1) + (iz))}{(2n+1)(2n)!}$$ What am I missing?

Answer 1

That equality is false. However, what is written at that link is not that. It's$$\sum_{n=0}^\infty\left(\frac{(iz)^{2n}}{(2n)!}+\frac{(iz)^{2n+1}}{(2n+1)!}\right)=\sum_{n=0}^\infty\frac{(iz)^n}{n!},\label{a}\tag1$$which is indeed true, since both series are equal to$$\frac1{0!}+\frac{iz}{1!}+\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\cdots$$More precisely, the LHS of \eqref{a} is what you get when you group in pairs the RHS of \eqref{a}, which is a legitimate choice, since we are dealing with absolutely convergent series here.