Laurent Series Of $\frac{z}{(z+1)(z+2)}$ Around $z=-1$

Question

Expand $\frac{z}{(z+1)(z+2)}$ Around $z=-1$

$$\frac{z}{(z+1)(z+2)}=-\frac{1}{z+1}+\frac{2}{z+2}=-(z+1)^{-1}+\frac{2}{z+2}=\\=-(z+1)^{-1}+\frac{2}{1-(-z-1)}=-(z+1)^{-1}+2\sum_{n=0}^{\infty}(-1)^n(z+1)^n$$

Can I write all inside the sum operator? the first terms are:

$\frac{-1}{(z+1)}+2-2(z+1)+2(z+1)^2+...$

So the function as a simple pole and the radius of convergence is $lim_{n\to \infty}\frac{1}{|\sqrt[n]{(-1)^n}|}=1$

So $|z+1|<1$ and on $|z+1|=1$ we get $\sum_{n=0}^{\infty}(-1)^n$ which diverges?

Answer 1

Perhaps would be simpler using $w=z+1$ then \begin{align} \frac{z}{(z+1)(z+2)} &=\frac{w-1}{w(w+1)}\\ &=\frac{w-1}{w}\frac{1}{1+w}\\ &=\frac{1-w}{-w}\left(1-w+w^2-w^3+\cdots\right)\\ &=\frac{1}{-w}\left(1-2w+2w^2-2w^3+\cdots\right)\\ &=-\frac{1}{w}+2-2w+2w^2-2w^3+\cdots\\ &=-\frac{1}{z+1}+2-2(z+1)+2(z+1)^2-2(z+1)^3+\cdots \end{align}

Answer 2

Your Laurent Series is correct. $$\frac{z}{(z+1)(z+2)}=-(z+1)^{-1}+2\sum_{n=0}^{\infty} (-1)^{n}(z+1)^n$$ You have to be careful about the region of convergence which is $0<|z+1|<1$