Find the Laurent series about $z_0 = 0$ for $f(z) = \frac{1}{z^5(1+z^3)}$ and show that it converges to $f(z)$ on the punctured disc {z ∈ C | 0 < |z| < 1}.
So I've gone and found what I think the Laurent series of f(z) as $\frac{1}{z^5} +\sum_{n=0}^{\infty} (-1)^{n+1} z^{3n-2}$ but i'm completely unsure of how to find the convergence and have it equal to f(z).
Transfer the terms with negative powers of $z$ to the left; what you have to show is that $\sum_{n=1} ^{\infty} (-1)^{n+1} z^{3n-2} =f(z)-\frac 1 {z^{5}} +\frac 1 {z^{2}}$. Some simple algebraic manipulations shows that $f(z)-\frac 1 {z^{5}} +\frac 1 {z^{2}}=\frac 1 {1+z^{3}}$. Now it is just a question of summing a simple geometric series.