What is the Laurent series representation of $f(z)=\frac{1}{z(1-z)(4-z)}$?
My solution: Using partial fractions, $f(z)=\frac{1}{4z} + \frac{1}{3(1-z)} - \frac{1}{12(4-z)}$
Hence, $$f(z)=\frac{1}{4} \sum_{n=0}^{\infty} (-1)^n (z-1)^n + \frac{1}{3} \sum_{n=0}^{\infty} z^n - \frac{1}{48} \sum_{n=0}^{\infty}\frac{z^n}{4^n}$$
Is this correct? And how can I put every term under just one summation?