I have given the following function $f(z)=\frac{z}{(z+1)(z+2)}$ and I need to find a power series expension in a nghd of $1$. I first remarked that $$f(z)=\frac{-1}{z+1}+\frac{2}{z+2}$$ But now I don't know how to find an expansion around $1$. Because we had a theorem that if $f$ is analytic then the expansion is given by the taylor series but here I first need to show that it is analytic which is exaclty finding an expansion.
Do you have a hint how I could start?
Best Answer
Use the fact that, if $|z-1|<3$,\begin{align}\frac2{z+2}&=\frac2{3+(z-1)}\\&=\frac23\frac1{1-(-(z-1)/3)}\\&=\frac23\sum_{n=0}^\infty\left(-\frac{z-1}3\right)^n\\&=\sum_{n=0}^\infty\frac{2(-1)^n}{3^{n+1}}(z-1)^n\end{align}and that, if $|z-1|<2$,\begin{align}\frac1{z+1}&=\frac1{2+(z-1)}\\&=\frac12\sum_{n=0}^\infty\frac{(-1)^n}{2^n}(z-1)^n\\&=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(z-1)^n\end{align}It follows that$$|z-1|<2\implies\frac z{(z+1)(z+2)}=\sum_{n=0}^\infty\left(\frac{2(-1)^n}{3^{n+1}}-\frac{(-1)^n}{2^n}\right)(z-1)^n$$