I'm trying to find a Power series (Laurent series), in a disk $B_1(0)$ , of $\frac{1}{(1-z)^m}$ where $m\in N$
2026-03-25 21:54:03.1774475643
Power series (Laurent series), in a disk $B_1(0)$ , of $\frac{1}{(1-z)^m}$ where $m\in N$
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If $m=1$, then you have$$\frac1{1-z}=1+z+z^2+z^3+\cdots.$$On the other hand,$$\frac1{(1-z)^2}=\left(\frac1{1-z}\right)'=1+2z+3z^2+4z^3+\cdots$$So$$\frac1{(1-z)^3}=\frac12\left(\frac1{(1-z)^2}\right)'=1+3z+6z^2+10z^3+\cdots$$Can you take it from here?