复变函数:f(z)=1/[z(z+1)];z=-1;展开洛朗级数

f(z)=1/[z(z+1)]=(1/z)-[1/(z+1)]
=[1/(z+1-1)]-[1/(z+1)]
=-[1/(1-(z+1))]-[1/(z+1)]
=-[1+(z+1)+(z+1)^2+(z+1)^3+……+(z+1)^n+……]-[1/(z+1)]
∞
=-∑(z+1)^k （z∈C,0