<s>[INST] Laurent series of $\frac{e^z}{z^2+1}$ 
I cant  figure out the laurent series of the following function.
Let $f(z)= \frac{e^z}{z^2+1} $ and $|z|\gt 1$
$$\frac{1}{z^2+1}=\sum_{n=0}^{\infty}(-1)^nz^{-2n-2}$$
and
$$e^z = \sum_{n=0}^{\infty}\frac{z^{n}}{n!}$$ 
$$e^z*\frac{1}{z^2+1} =\sum_{n=0}^{\infty}\sum_{k=0}^{n}(-1)^nz^{-2n-2}*\frac{z^{n-k}}{(n-k)!}$$
How can I go from here ?
 [/INST] 
Let
\begin{align*}
 f(z)&= \frac{e^z}{z^2+1}
\end{align*}

We assume we need a Laurent expansion for all $|z|>1$  with center $z_0=0$.

We    obtain
  \begin{align*}
f(z)&=\frac{e^z}{z^2+1}\\
&=\frac{e^z}{z^2}\cdot\frac{1}{1+\frac{1}{z^2}}\\
&=\frac{e^z}{z^2}\sum_{k=0}^\infty(-1)^k\frac{1}{z^{2k}}\\
&=\sum_{j=0}^\infty\frac{z^j}{j!}\sum_{k=0}^\infty(-1)^k\frac{1}{z^{2k+2}}\\
&=\sum_{n=-\infty}^\infty\left(\sum_{{j-2k-2=n}\atop{j\geq 0;k\geq  0}}\frac{(-1)^k}{j!}\right)z^n\tag{1}\\
&=\sum_{n=-\infty}^\infty\left(\sum_{{j-2k=n}\atop{j\geq 0;k\geq  1}}\frac{(-1)^{k+1}}{j!}\right)z^n\tag{2}\\
&=\sum_{n=-\infty}^\infty\left(\sum_{k=K}^\infty\frac{(-1)^{k+1}}{(n+2k)!}\right)z^n\tag{3}\\
\end{align*}

Comment:


*

*In (1) we apply the Cauchy product formula

*In (2) we shift the index $k$ by one

*In (3) we replace $j$ with $n+2k$ and since $j=n+2k\geq 0$ and $k\geq 1$ we have to set
\begin{align*}
K=\max\left\{1,-\left\lfloor\frac{n}{2}\right\rfloor\right\}
\end{align*}
</s>