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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?

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I think you're right to be leery; 'Maclaurin' generally refers specifically to Taylor series centered around $z=0$, and Taylor itself implies no negative terms (otherwise it would be a Laurent series). – Steven Stadnicki May 8 '13 at 22:40
do you think it makes this expansion any less valid? – franklin May 8 '13 at 22:41
Instead, I would consider the Maclaurin series for $\exp(z)$ and for $\dfrac1{z+1}$ separately - both of which should be very easy to find - and then multiply them. – Steven Stadnicki May 8 '13 at 22:42
I think your expansion is valid, but I don't think it's the expansion that the question is looking for. – Steven Stadnicki May 8 '13 at 22:42
What you found is the Laurent expansion around $z = -1$. – Random Variable May 8 '13 at 23:05

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 $$\frac{e^z}{z+1}=\sum_{n=0}^\infty\frac{z^n}{n!}\times\sum_{n=0}^\infty(-z)^n=\sum_{n=0}^\infty c_nz^n$$

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@StevenStadnicki I think now I have a correct answer. Isn't it? – user63181 May 8 '13 at 23:11
Is $c_n$ intended to represent some well known sequence, or are you just defining $c_n=\sum_{k=0}^n \frac{(-1)^k}{k!}$? – Managu May 9 '13 at 0:54
@SamiBenRomdhane It looks essentially correct to me; I would've phrased it differently, but all the maths are correct. – Steven Stadnicki May 9 '13 at 3:07

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