<s>[INST] Determine the Maclaurin series expansion of $\frac{\mathrm{exp}(z)}{(z+1)}$ Determine the Maclaurin series expansion of $\frac{\mathrm{exp}(z)}{(z+1)}$. 
This is the composition of the series expansion of the exponential function centered about $z = -1$. We can rectify the expansion about $\mathrm{exp}(z)$ by writing:
$$\begin{aligned} 
\frac{\mathrm{exp}(z + 1)}{e (z+1)} &= \frac{1}{e(1+z)} \sum_{n=0}^\infty \frac{(z+1)^n}{n!} \\ 
&= \frac{1}{e} \sum_{n=0}^\infty \frac{(z+1)^{(n-1)}}{n!} 
\end{aligned}$$ 
Is this expansion correct? There is a slight uneasiness here that I have with calling this a Maclaurin expansion due to the centering and the negative power when $n = 0$.
Any thoughts?
 [/INST] The function $e^z$ is entire and has Maclaurin series:
$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!} + \cdots$$
The function $\frac{1}{1+z}$ has a pole at $z=-1$ and has a Maclaurin series which converges for $|z|<1$:
$$\frac{1}{1+z}=1-z+z^2-z^3+ \cdots$$
Their product has a series that converges for $|z|<1$:
$$\frac{e^z}{1+z} = 1+\frac{z^2}{2} -\frac{z^3}{3}+ \frac{3z^4}{8} -\frac{11z^5}{30}+ \frac{53z^6}{144} - \cdots$$
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