How to find power series of $f(z)=\frac{e^z}{1-z}$ at $z_0=0$? I tried to calculate few derivatives, but I cant get $f^{(n)}(z)$ from them. Any other way?   
$$f(z)=\frac{e^z}{1-z}\text{ at }z_0=0$$
 A: Since, formally,
$$ \frac{1}{1-z}=1+z+z^2+z^3+\ldots $$
the multiplication by $\frac{1}{1-z}$ brings the power series $a_0 + a_1 z+ a_2 z^2 +\ldots $ into the power series $a_0+(a_0+a_1)z+(a_0+a_1+a_2) z^2+\ldots$. It follows that:
$$ \frac{e^{z}}{1-z}=\sum_{n\geq 0}z^n \left(\sum_{j=0}^{n}\frac{1}{j!}\right).$$
A: Hint:
$$\frac1{1-z}=\sum_{n=0}^\infty z^n$$
$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$
Now apply Cauchy products to see that
$$\frac{e^z}{1-z}=\sum_{n=0}^\infty z^n\sum_{k=0}^n\frac1{k!}=\sum_{n=0}^\infty e_n(1)z^n$$
where $e_n(x)$ is the exponential sum formula.
A: Standard fact the coefficient for the series $\frac{f(z)}{1-z}$ is $\sum_{n=0}^{\infty}(\sum_{k=0}^{n}a_k)x^n$ where $f(z)$ has the expansion $\sum_{n=0}^{\infty}a_nx^n$
A: $$
g(z) = a_0+a_1z+a_2z^2+...,\\
\frac{g(z)}{1-z}=a_0\frac{1}{1-z}+a_1\frac{z}{1-z}+a_2\frac{z^2}{1-z}+...
$$
Using the power series for $\frac{1}{1-z}$ gives
$$
g(z)=a_0(1+z+z^2+...)+a_1z\cdot(1+z+z^2+...)+a_2z^2\cdot(1+z+z^2+...),\\
g(z)=a_0(1+z+z^2+...)+a_1\cdot(z+z^2+z^3+...)+a_2\cdot(z^2+z^3+z^4...),\\
g(z)=a_0+(a_0+a_1)z+(a_0+a_1+a_2)z^2+...
$$
For the specific case $g(z)=e^z$
$$
\frac{e^z}{1-z}=\frac{1}{0!}+\left(\frac{1}{0!}+\frac{1}{1!}\right)z+\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}\right)z^2+...+\left(\sum_{r=0}^n\frac{1}{r!}\right)z^n+...
$$
as required.
