calulate the Fouries Series of $e^{2cosx}$. I don't understand the calculate process. $\sum_{m=0}^{\infty}\frac{z^m}{m!}\sum_{l=0}^{\infty}\frac{z^{-l}}{l!} = \sum_{n=0}^{\infty}\sum_{m=0}^n \frac{z^m}{n!(n-m)!}$ I stop in this step. So how the $\sum_{n=0}^{\infty}\sum_{m=0}^n \frac{z^m}{n!(n-m)!}$ can calulate the last result?
Let $n:=m-l$, then we observe at first \begin{align*} \sum_{m \in \mathbb N_0}\frac {z^m}{m!} \sum_{l\in \mathbb N_0}\frac{z^{-l}}{l!} &=\sum_{m \in \mathbb N_0} \sum_{l\in \mathbb N_0}\frac{z^{m-l}}{m!l!} = \sum_{\substack{m,l \in \mathbb N_0}\\ n=m-l}\frac{z^{n}}{m!(m-n)!}. \end{align*} We want to rewrite the sums in terms of $n$. For $m,l\in \mathbb N_0$ we know that $n$ can take any value in $\mathbb Z$. Hence, the right hand side can be rewritten as $$\sum_{\substack{n\in \mathbb Z\\m,l\in \mathbb N_0\\n=m-l}}\frac{z^{n}}{m!(m-n)!}=\sum_{n\in\mathbb Z}\sum_{\substack{m,l\in \mathbb N_0\\n=m-l}}\frac{z^{n}}{m!(m-n)!}.$$ For some fixed $n\in \mathbb Z$ let $m,l \in \mathbb N_0$ be a solution of $n=m-l$, then we have $m\geq n$. Moreover, let $m\geq n$, then m,l is a solution of $n=m-$ if we choose $l:=m-n$. In other words we can rewirte the right sum by indexing over $m\geq n$, so the sum above becomes $$\sum_{n\in\mathbb Z}\sum_{m\geq n}\frac{z^{n}}{m!(m-n)!}=\sum_{n\in \mathbb Z}\left ( \sum_{m\geq n} \frac 1 {m!(m-n)!} \right)z^n.$$