Laurent series $\ \ \ \ \ $ Trying to compute first few terms of the Laurent series for:
$$ \frac{e^z}{z^2(z^2+1)} =\sum_{n=-2}^\infty c_n z^n = \quad ?$$
I know the expansion of $$e^z = \sum_{n=0}^\infty  \frac{z^n}{n!}$$  and I could have $\frac{e^z}{z^2}$ expanded. But how do I fit all together and be able to see when it converges.
From the comment I have:
$$ \frac{e^z}{1+z^2} = \sum_{n=0}^\infty  \frac{z^n}{n!}
\sum_{n=0}^\infty (-1)^n z^{2n} 
 $$
And using Caucy product: $$ c_n =\sum_{k=0}^n  \frac{z^k}{k!} (-1)^{n-k} z^{2(n-k)} $$
It follows that:
$$ \frac{e^z}{z^2(z^2+1)} = \frac{1}{z^2}\sum_{n=0}^\infty \sum_{k=0}^n  \frac{z^k}{k!} (-1)^{n-k} z^{2(n-k)}  $$

Via the Cauchy product,
\begin{align}
\frac{e^z}{z^2+1} 
&= e^z\cdot\frac{1}{1-(-z^2)} \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (-z^2)^n\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (iz)^{2n}\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{1+(-1)^n}{2}(iz)^n\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{i^n+(-i)^n}{2}z^n\right) \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{1}{k!} \cdot\frac{i^{n-k}+(-i)^{n-k}}{2}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{i^{n-k}+(-i)^{n-k}}{2(k!)}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{i^k+(-i)^k}{2((n-k)!)}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{1+(-1)^k}{2}\cdot\frac{i^k}{(n-k)!}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{i^{2k}}{(n-2k)!}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^n. \tag1
\end{align}
So the desired Laurent series is
\begin{align}
\frac{e^z}{z^2(z^2+1)} 
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^{n-2} \\ 
&= \sum_{n=-2}^\infty \left(\sum_{k=0}^{(n+2)/2} \frac{(-1)^k}{(n+2-2k)!}\right) z^n. 
\end{align}

In hindsight, I took the long way to obtain $(1)$.  Here's a shorter approach:
\begin{align}
\frac{e^z}{z^2+1} 
&= \frac{1}{1-(-z^2)}\cdot e^z \\
&= 
\left(\sum_{n=0}^\infty (-z^2)^n\right) 
\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)
\\
&= 
\left(\sum_{n=0}^\infty (iz)^{2n}\right) 
\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)
\\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} i^{2k}\cdot \frac{1}{(n-2k)!}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^n.
\end{align}

