Find the coefficient of $z$ in the Laurent series expansion of $\frac{e^z}{z-1}$ in ${|z| > 1}$

Question

I've found this duplicate from '15: Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$, but I think it's wrong, since it looks for the Laurent expansion in ${|z-1|>1}$.

After developing it by myself, I've reached the following:

$$f(z)=\frac{e^z}{z-1}=\frac{1}{z-1}\cdot e^z$$

Then, $e^z=\sum_{n\geq 0} \frac{z^n}{n!}$ and $$\frac{1}{z-1}=\sum_{n \geq 0} \frac{1}{z^{n+1}}$$.

Here I thought of using the Cauchy Product, but I end up with

$$f(z)=\sum_{n \geq 0} \sum_{k=0}^n \frac{z^{2k-n-1}}{k!}$$ (How can I make this font bigger? The exponent is too small. Sorry about that. The exponent would be $2k-n-1$).

So I have to sum whenever $2k-n-1 = 1$

If I'm not getting it wrong, this means that I have to calculate $$\sum_{n \geq 0} \frac{1}{(1+2n)!}$$ (I'm not sure how to show it better, but I've put the first 3 terms and were $\frac{1}{1!}, \frac{1}{3!}$ and $\frac{1}{5!}$. I'm not sure how to calculate that. The nearest thing I thought was that $$\sum_{n \geq 0} \frac{1}{n!}=e $$. Maybe what I found before could be considered as an answer, but I think I can go one step further.

Thanks

Answer 1

We have

$$\frac{e^z}{z-1} = e^z \cdot \frac{1}z \cdot \frac1{1-\frac1z} = \left(\sum_{n=0}^\infty \frac{z^{n}}{n!}\right)\left(\sum_{n=0}^\infty \frac{1}{z^{n+1}}\right)$$

so the coefficient of $z$ is equal to

$$\sum_{n=2}^\infty \frac1{n!} = e-2$$

It is because $z$ is obtained when multiplying the term $\frac{z^{k+2}}{(k+2)!}$ from the first sum, with the term $\frac1{z^{k+1}}$ from the second sum:

$$\frac{z^2}{2!}\cdot \frac1z + \frac{z^3}{3!}\cdot \frac1{z^2} + \frac{z^4}{4!}\cdot \frac1{z^3} + \cdots = z \left(\frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \cdots \right)$$

Answer 2

We denote with $[z^n]$ the coefficient of $z^n$ of a series.

We obtain for $n\in\mathbb{Z}$ \begin{align*} [z^n]\frac{e^z}{z-1}&=[z^{n}]\frac{e^z}{z}\cdot\frac{1}{1-\frac{1}{z}}\tag{1}\\ &=[z^{n+1}]\sum_{k=0}^\infty\frac{z^k}{k!}\sum_{j=0}^\infty\frac{1}{z^j}\tag{2}\\ &=\sum_{k=0}^\infty \frac{1}{k!}[z^{n+1-k}]\sum_{j=0}^\infty\frac{1}{z^j}\tag{3}\\ &=\left\{ \begin{array}{ll} \sum_{k=n+1}^\infty\frac{1}{k!}&\qquad n\geq 0\\ \sum_{k=0}^\infty\frac{1}{k!}&\qquad n< 0\\ \end{array} \right.\tag{4}\\ &\color{blue}{=}\left\{ \begin{array}{ll} \color{blue}{e-\sum_{k=0}^n\frac{1}{k!}}&\color{blue}{\quad\, n\geq 0}\\ \color{blue}{e}&\color{blue}{\quad\, n< 0}\\ \end{array} \right. \end{align*}

Comment:

• In (1) we factor out $z$.

• In (2) and (3) we use $[z^p]z^qA(z)=[z^{p-q}]A(z)$.

• In (4) we select the coefficient of $z^{n+1-k}$.

We conclude the Laurent expansion of $\frac{e^z}{z-1}$ in $|z|>1$ is \begin{align*} \color{blue}{\frac{e^z}{z-1}=e\sum_{n=-\infty}^{-1}z^n + \sum_{n=0}^\infty\left(e-\sum_{k=0}^n\frac{1}{k!}\right)z^n} \end{align*}