Trying to calculate the series $\sum_{n=0}^{\infty}{{-1}\choose n}z^n$

Question

I'm trying to calculate the series $\sum_{n=0}^{\infty}\binom{-1}{n}z^n$.

Here is what I have so far:

\begin{align*} \sum_{n=0}^{\infty}\binom{-1}{n}z^n &= \sum_{n=0}^{\infty}\bigg(\frac{1}{n!}\cdot\prod_{j=0}^{n-1}(-1-j)\bigg)z^n \\&= \sum_{n=0}^{\infty}\bigg(\frac{1}{n!}\cdot (-1)^n\cdot\prod_{j=0}^{n-1}(j+1)\bigg)z^n \\&= \sum_{n=0}^{\infty}\bigg(\frac{1}{n!}\cdot n! \cdot (-1)^n\bigg)z^n \\&= \sum_{n=0}^{\infty}(-1)^nz^n \end{align*} Now I am kind of stuck. Trying to apply the Cauchy product I just end up with $$\sum_{n=0}^{\infty}(-1)^nz^n=\sum_{n=0}^{\infty}\bigg(\sum_{k=0}^{n}(-1)^k\cdot z^k\cdot z^{n-k}\bigg)=\bigg(\sum_{n=0}^{\infty}(-1)^nz^n\bigg) \cdot \bigg(\sum_{n=0}^{\infty}z^n\bigg)$$ leading to some sort of infinite regress. Anybody got a clue for me?

Answer 1

What you have so far is correct. What's missing in the last step is to apply the geometric series formula $$ \sum_{n=1}^\infty ar^n = a + ar + ar^2 + \cdots = \frac{a}{1-r},$$ which is valid for $|r|<1$. Have you seen this formula before?

(If you are interested in a nice generalization of this identity, check this out.)

Answer 2

Note that $$ \binom{-1}{n}=(-1)^{n}\frac{n!}{n!}=(-1)^n. $$ Hence $$ \sum_{n=0}^\infty \binom{-1}{n}z^{n} =\sum_{n=0}^\infty(-z)^n=\frac{1}{1+z};\quad (|z|<1) $$ by the geometric series.