裂项求和法律法规
先裂项
f(z)=z/(z+1)(z+2)=-1/(1+z)+2/(2+z)
再根据需要变项
f(z)=-1/(3+z-2)+2/(4+z-2)=(-1/3){1/[1-[(-1)(z-2)/3]}+(1/2){1/[1-[(-1)(z-2)/4]}
再展开,z-2的绝对值小于3(因为-1是奇点)
f(z)=sigma{(-1/3)[(-1)(z-2)/3]^n}+(1/2)[(-1)(z-2)/4]^n}
sigma是求和号,从n=0到无穷
先裂项
f(z)=z/(z+1)(z+2)=-1/(1+z)+2/(2+z)
再根据需要变项
f(z)=-1/(3+z-2)+2/(4+z-2)=(-1/3){1/[1-[(-1)(z-2)/3]}+(1/2){1/[1-[(-1)(z-2)/4]}
再展开,z-2的绝对值小于3(因为-1是奇点)
f(z)=sigma{(-1/3)[(-1)(z-2)/3]^n}+(1/2)[(-1)(z-2)/4]^n}
sigma是求和号,从n=0到无穷