Find the Laurent expansion for $$\frac{1}{z-1}\frac{1}{z+3}$$ inside the annulus $1<\vert z\vert<3$.
Since $|z|>1$, $$\frac{1}{z-1}=\frac{-1}{1-z}=\frac{-1}{(1-1/z)(-z)}=\frac{1}{z}\sum_{n=0}^\infty z^{-n}=\sum_{n=0}^\infty z^{-n-1}.$$
Since $\vert z \vert <3,$ we have $$\frac{1}{z+3}=\frac{1}{3(1-(-z/3))}=\sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}z^n.$$
How can the product of these two series be simplified into a Laurent series? I have tried using the formula $$\Bigg(\sum_{n=0}a_n\Bigg)\Bigg(\sum_{n=0}b_n\Bigg) = \sum_{n=0}\Bigg(\sum_{m=0}^n a_mb_{n-m}\Bigg)$$ with $a_n=z^{-n-1}$ and $b_n=\frac{(-1)^n}{3^{n+1}}z^n$, but that leads to $$\sum_{n=0}\bigg(\sum_{m=0}^n z^{-m-1}z^{n-m}(-1)^{n-m}/3^{n-m-1}\bigg)=\sum_{n=0}\bigg(\sum_{m=0}^n z^{n-2m-1}(-1)^{n-m}/3^{n-m-1}\bigg)$$
Partial fractions: $$ \dfrac{1}{(z-1)(z+3)} = \dfrac{1}{4(z-1)} - \dfrac{1}{4(z+3)}$$