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?
Inspired by the comment of Steven Stadnicki, the Maclaurin series is a taylor series centred at the origin so by the Cauchy product and since
$$\sum_{k=0}^n \frac{z^k}{k!}\times(-z)^{n-k}=\sum_{k=0}^n (-1)^k\frac{z^n}{k!}=z^n\sum_{k=0}^n\frac{(-1)^k}{k!}=c_nz^n$$ we find $$\frac{e^z}{z+1}=\sum_{n=0}^\infty\frac{z^n}{n!}\times\sum_{n=0}^\infty(-z)^n=\sum_{n=0}^\infty c_nz^n$$