TITLE: Power series problem in complex analysis
QUESTION [5 upvotes]: Suppose that $f(z)= (e^z)/(1-z)$ 
How can I find out Power series expansion of f about $z=0$??
Is the use of cauchy product must here ? Can it be done without using cauchy product?Please help..
Many many thanks in advance.

REPLY [4 votes]: Hint: It's useful to know that multiplication with $\frac{1}{1-z}$ means to sum up the coefficients.

Let $A(z)=\sum_{n=0}^{\infty}a_nz^n$ be a power series. We obtain when multiplying with $\frac{1}{1-z}$
  \begin{align*}
A(z)\frac{1}{1-z}&=\left(\sum_{k=0}^{\infty}a_zz^k\right)\left(\sum_{l=0}^{\infty}z^l\right)\\
&=\sum_{n=0}^{\infty}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}a_k\right)z^n\\
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^na_k\right)z^n\\
\end{align*}

So, whenever we need the sum $\sum_{k=0}^{n}a_k$ of coefficients $a_k$ we simply have to multiply the corr. power series with $\frac{1}{1-z}$.

Keeping this in mind, you may conclude immediately
  \begin{align*}
\frac{e^z}{1-z}=\frac{1}{1-z}\sum_{n=0}^\infty\frac{z^n}{n!}
=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\frac{1}{k!}\right)z^n
\end{align*}