Write f(z) as a power series around zero and identify the radius of convergence $f(z) = \frac{e^z}{z-1}$

Question

Consider the function f : $\mathbb{C}$ → $\mathbb{C}$ given by

$f(z) = \frac{e^z}{z-1}$.

a) Using Taylor’s formula or otherwise, write f as a power series around zero. That is, find $a_0, a_1, . . .$ in $\mathbb{C}$ such that

$f(z) = \sum_{n=0}^{\infty} a_n z^n$.

b) Identify the radius of convergence of the power series. Give reasoning for your answer.

Really struggling with this, not familiar with the power series $\frac{1}{z-1}$, have read somewhere that $\frac{1}{z-1}$ = $-\frac{1}{1-z}$ and I know $\frac{1}{1-z}$$=\sum_{n=0}^{\infty}z^n$ and $e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$ just unsure how to write $f(z) = \frac{e^z}{z-1}$ as a power series with the knowledge I have. Also any help with part (b) would be great too!

Answer 1

hint

$$(z-1)f(z) = e^z$$

$$\sum_{n=1}^\infty a_{n-1}z^{n}-\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty \frac{z^n}{n!}$$

then

$$a_0=1$$ and for $ n\ge 1,$

$$a_n=a_{n-1}-\frac{1}{n!}$$

Use Telescoping to find $ a_n$.

Answer 2

By Cauchy product it's simple to find : $$f(z)=\sum_{n=0}^\infty \biggl(\sum_{k=0}^n\frac{-1}{k!}\biggr)z^n$$