Originally Posted by

**arbolis** I made a mistake, I've edited my post. However I still don't reach the Taylor series you provided. I don't see why the series I found is not a Taylor series.

I've used the fact that $\displaystyle f(z)=\sum _{n=0}^{\infty} \frac{f^{(n)}(z-z_0)}{n!}$.

I've found out that $\displaystyle f^{(n)}=\frac{(-1)^n (n+1)!}{z^{(n+2)}}$.

Now you make me doubt the formula holds for the function $\displaystyle z \mapsto \frac{1}{z^2}$ due to the singularity at $\displaystyle z=0$.

Edit : I understand the equality you gave, but I don't know how to reach it. I mean I don't know how to reach that the Taylor Series is $\displaystyle 1 + 2(z + 1) + 3(z + 1)^2 + 4(z + 1)^3 + ...$

Edit 2 : Sorry, I just found my error. I forgot to evaluate $\displaystyle f^{(n)}$ in $\displaystyle z_0$...

I think I'll do it. Thanks a lot!