Laurent Series Expansion and Convergence Proof

Question

Show that for any integer $k ≥ 0$ we have

$$\frac{1}{(1-z)^{k+1}} = \sum\limits_{n=0}^{\infty}\binom {n+k}{n}z^{n}$$

and that it converges absolutely for any $|z| < 1$.

I really do not just know how to approach this question.

Answer 1

You might try induction, noting that $$ \dfrac{d}{dz} \frac{1}{(1-z)^{k+1}} = \frac{k+1}{(1-z)^{k+2}}$$

Answer 2

The title of your question is strange, since there's no singularity here and therefore what you call a Laurent series is just a Taylor series.

If $f(z)=(1-z)^{-k-1}$, then$$(\forall n\in\mathbb{N}):f^{(n)}(z)=(k+1)(k+1)\cdots(k+n)(1-z)^{-k-n-1}$$and therefore$$(\forall n\in\mathbb{N}):\frac{f^{(n)}(0)}{n!}=\frac{(k+1)(k+1)\cdots(k+n)}{n!}=\binom{n+k}n.$$You can prove (as it was suggested in the comments) that the series$$\sum_{n=0}^\infty\binom{n+k}nz^n\tag1$$converges when $|z|<1$ through that ratio test and then a standard Complex Analysis theorem tells us that the sum of the series $(1)$ is $f(z)$ in that region.