I need to find Laurent expansion for $$\frac{e^{\frac{1}{z-1}}}{z(z+1)}$$ about $z=1$ for $1 \lt |z-1| \lt 2$.
I've tried to solve this as it was suggested in the book: find Laurent expansion for $f_1(z) = \frac{1}{z+1}$ about $z=1$ for $|z-1| \lt 2$ and Taylor expansion for $f_2(z) = \frac{e^{\frac{1}{z-1}}}{z}$ for $|z-1| > 1$ (this function is analytic in this area, so Taylor series is equal to Laurent series) and then multiply them.
I get $\frac{1}{z+1} = \sum_{n=-\infty}^{-1}(-1)^{n-1}2^{-n-1}(z-1)^n$, $\frac{1}{z} = \sum_{n=0}^{\infty}(-1)^n(z-1)^n$ and $e^{\frac{1}{z-1}} = 1+ \sum_{n=1}^{\infty}\frac{1}{n!(z-1)^n}$.
But I don't understand how can I find product of these series?
Thanks for your help.