Find the series representation of the function $f(z)=\frac{\exp {z}}{1-z}$

Question

Find the series representation of the function $f(z)=\frac{\exp {z}}{1-z}$ with $z\in\mathbb{C}$

My attempt: Let $$g(z)=\exp(z)=\sum_{n\geq0}\frac{z^n}{n!}$$

$$h(z)=\frac{1}{1-z}=\sum_{n=0}^\infty z^n$$

Then

$$f(z)=\frac{\exp {z}}{1-z}=\sum_{n=0}^\infty c_nz^n$$

Where $$c_n=\sum_{k=0}^n\frac{1}{k!}$$

Here i'm stuck. Can someone help me?

Answer 1

Using induction it isn't hard to prove that for any $g \in C^{\infty}(z)$ then $$\frac{\partial^n}{\partial z^n} (e^zg(z))=e^z\sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z) $$ In particular, $g(z)=\frac{1}{1-z}$ one has $\frac{\partial^k g}{\partial z^k}(z)=\frac{k!}{(1-z)^{k+1}}$ so the series becomes $$f(z)=\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)k!\right)\frac{z^n}{n!}}$$

On edit:

For the induction step, let $\frac{\partial^n}{\partial z^n} (e^zg(z))=e^z\sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z) $ which is obvious for $n=0$.

Now assume is true for $n$ and compute for $n+1$: It becomes $\frac{\partial^{n+1}}{\partial z^{n+1}} (e^zg(z))=\frac{\partial}{\partial z}\left(\frac{\partial^n}{\partial z^n} (e^zg(z))\right)$ $=\frac{\partial^n}{\partial z^n} (e^zg(z))+e^z\frac{\partial}{\partial z}\left(\sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z)\right) = \frac{\partial^n}{\partial z^n} (e^zg(z))+e^z\left(\sum_{k=1}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{ \partial^{k+1} g}{\partial z^{k+1}}(z)\right) = e^z\left(\sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z) + \sum_{k=0}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{ \partial^{k+1} g}{\partial z^{k+1}}(z)\right) = e^z\left(g(z) + \sum_{k=1}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z) + \sum_{k=0}^{n-1}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{ \partial^{k+1} g}{\partial z^{k+1}}(z) + \frac{ \partial^{n+1} g}{\partial z^{n+1}}(z) \right) = e^z\left(g(z) + \sum_{k=1}^{n}\left(\begin{array}{c} n \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z) + \sum_{j=1}^{n}\left(\begin{array}{c} n \\ j-1 \end{array} \right)\frac{ \partial^{j} g}{\partial z^{j}}(z) + \frac{ \partial^{n+1} g}{\partial z^{n+1}}(z) \right) = e^z\left(g(z) + \sum_{k=1}^{n}\left(\left(\begin{array}{c} n \\ k \end{array} \right)+\left(\begin{array}{c} n \\ k-1 \end{array} \right)\right)\frac{\partial^k g}{\partial z^k}(z) + \frac{ \partial^{n+1} g}{\partial z^{n+1}}(z) \right) = e^z\left(g(z) + \sum_{k=1}^{n}\left(\begin{array}{c} n+1 \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z) + \frac{ \partial^{n+1} g}{\partial z^{n+1}}(z) \right) =e^z\sum_{k=0}^{n+1}\left(\begin{array}{c} n+1 \\ k \end{array} \right)\frac{\partial^k g}{\partial z^k}(z) $.

The claim holds true.

Answer 2

$(1+z + z^2 + \cdots)(1 + z + \frac12 z^2 + \frac 1{3!} z^3 + \cdots)$

What is the coefficient of the $z^n$ term? We can ignore all of the terms that are greater than $z^n.$ Take the first term of the factor on the left, multiply by the corresponding term on the factor on the right. Then move the the next pair of terms.

$(1)(\frac 1{n!} z^n) + (z)(\frac{1}{(n-1)!} z^{n-1}) +\cdots + (z^k)( \frac {1}{n-k} z^k) + \cdots + (z^n)\cdot (1) = \sum_\limits{k=0}^n \frac {1}{k!}$

Answer 3

Compute the first terms to get $$f(z)=\frac{e^z}{1-z}=1+2 z+\frac{5 z^2}{2}+\frac{8 z^3}{3}+\frac{65 z^4}{24}+\frac{163 z^5}{60}+\frac{1957 z^6}{720}+O\left(z^7\right)$$ In $OEIS$ you will find the numerator and denominator of the coefficients in sequences $A061354$ and $A061355$. $$f(z)=\sum_{n=0}^\infty \frac{e\, \Gamma (n+1,1)}{n!}x^n$$