let define : Laurent series of the function $f(z)=e^{z+\frac{1}{z}}$ around zero:
$$\sum^{\infty}_{n=-\infty}a_n z^n$$
prove that for all $n\in\mathbb{N}$ :
$$a_n=a_{-n}=\sum^{\infty}_{k=0}\frac{1}{k!(n+k)!}.$$
I did the following steps :
$$e^z=\sum^{\infty}_{n=0}\frac{z^n}{n!},$$
$$e^{z+\frac{1}{z}}=\sum^{\infty}_{n=0}\frac{({z+\frac{1}{z}})^n}{n!}.$$
I used the Binomial theorem:
$$\frac{({z+\frac{1}{z}})^n}{n!}=\sum^{n}_{k=0}\frac{{{n}\choose{k}}z^k\cdot \frac{1}{z}^{n-k}}{n!}=$$ $$=\sum^{n}_{k=0}\frac{z^{2k-n}}{k!(n-k)!}.$$
So the Laurent series is :
$$e^{z+\frac{1}{z}}=\sum^{\infty}_{n=0}\sum^{n}_{k=0}\frac{z^{2k-n}}{k!(n-k)!}.$$ How to proceed from here ?