Power series problem in complex analysis 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.
 A: Your $f(z)$ is the product of two functions with explicit power series expansion at $z=0$. Namely
$$e^z = \sum_{n=0}^\infty \frac{z^n}{n!},$$
and
$$\frac{1}{1-z} = \sum_{m=0}^\infty z^m.$$
Thus 
$$f(z) = e^z \cdot \frac{1}{1-z} =  \sum_{n=0}^\infty \frac{z^n}{n!} \cdot \sum_{m=0}^\infty z^m = \sum_{j=0}^\infty z^j \left(\sum_{l=0}^j \frac{1}{l!} \right).$$
I actually don't think that the sum $\sum_{l=0}^j \frac{1}{l!}$, as you say obtained by Cauchy product, can be further simplified (except that it converges to $e$ for $j\rightarrow \infty$), but I could be mistaken.
A: Cauchy product is the quickest way to do this.
$$\frac{1}{1-z}=\sum_n z^n$$
$$e^z=\sum_n \frac{z^n}{n!}$$
$$\frac{e^z}{1-z}=\sum_{n=0}^\infty z^n\sum_{k=0}^n \frac{1}{k!}=\sum_{n=0}^\infty\frac{z^n}{n!}\sum_{k=0}^n \frac{n!}{k!}$$
The coefficients are $\sum_{k=0}^n \frac{n!}{k!}$ are integers that satisfy the recursion formula $F_n=n F_{n-1}+1$, $F_0=1$.
Wolfram Alpha suggests this sum equals $e \Gamma(n+1,1)$ (incomplete Gamma function).
A: Your function satisfies the differential equation
$$
f'(z)=f(z) \ g(z)
$$
where
$$
g(z)= \frac{2-z}{1-z} = 1+\frac{1}{1-z} = 2+z+z^2+z^3+z^4+z^5+\cdots
$$
We can apply Leibniz rule to $f'(z)=f(z) \ g(z)$ and compute all derivatives of $f$ at $z=0$ recursively, since the derivatives of $g$ at $z=0$ are known:
$$
f^{(n+1)}(0)=\sum_{k=0}^n {n \choose k} f^{(k)}(0) g^{(n-k)}(0)
$$
The coefficients of the expansion of $f$ around $z=0$ are of course $\dfrac{f^{(n)}(0)}{n!}$.
This solution does not use Cauchy product but is probably more work.
A: 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*}

A: Using the cauchy product is a good idea:
We know 
$$e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$
and
$$\frac{1}{1-z} = \sum_{k=0}^\infty z^k \text{ for } |z| < 1.$$
So, by the cauchy product, we have around $z = 0$:
$$f(z) = \left( \sum_{k=0}^\infty \frac{z^k}{k!} \right) \left(\sum_{k=0}^\infty z^k \right) = \sum_{k=0}^\infty \sum_{l=0}^k \left( \frac{1}{l!} \cdot 1 \right) z^k = \sum_{k=0}^\infty \sum_{l=0}^k \frac{1}{l!} z^k$$
I'm not aware of a simpler expression for $\sum_{l=0}^k \frac{1}{l!}$, although there might be. 
You can also try to find a general expression for the $k$-th derivative of $f$,
and then $f^{(k)}(0)/k!$ gives the $k$-th coefficient of the power series of $f$ around zero. Let's take a look at the first derivatives:
$$f^{(1)}(z) = - \frac{e^z (z-2)}{(1-z)^2}$$
$$f^{(2)}(z) = \frac{e^z (z^2-4z+5)}{(1-z)^3}$$
$$f^{(3)}(z) = - \frac{e^z (z^3-6z^2+15z-16)}{(1-z)^4}$$
Therefore, the general form should look something like
$$f^{(n)}(z) = (-1)^n \frac{p_n(z) e^z }{(1-z)^{n+1}}$$
where $p_n(z)$ is some polynomial of degree $n$. But we need the explicit form of $p_n(z)$, since the constant term changes the value of $f^{(n)}(0)$. This seems like a significantly harder problem than using the cauchy product, so I would advise against it.
