Hi guys I am trying to obtain the power series of $f(z)=\frac{e^z}{(1-z)}$.
My idea was that I know $e^z= \sum \frac{z^n}{n!}$ and $\frac{1}{1-z}=\sum z^n$ for $|z|<1$. That is good so far, but now I am stuck
$f(z)= e^z* \frac{1}{1-z}=\sum \frac{z^n}{n!} * \sum z^n$
I am not sure how to multiply those. Is this some binomial type of calculation I should recognize or there there is a better way of approach.
For $\lvert z \rvert \lt 1,$ both series involved are absolutely convergent, so you can just multiply them term by term, arranging the terms as needed:
\begin{align} \frac{e^z}{1-z}&=\sum_{j=0}^\infty \frac{z^j}{j!}\cdot\sum_{k=0}^\infty z^k \\&=\sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{j+k}}{j!} \\&=\sum_{n=0}^\infty \sum_{\substack{j,\,k\,\ge\, 0\\j+k\,=\,n}}\frac{z^n}{j!} \\&=\sum_{n=0}^\infty\sum_{j=0}^n \frac{z^n}{j!} \\&=\sum_{n=0}^\infty\big(\sum_{j=0}^n \frac1{j!}\big)\, z^n. \end{align}