Prove that $\sum _{n=0}^{\infty} z^n = \dfrac{1}{1-z}$, if $|z|<1$

Question

Here $z \in \mathbb{C}$. If $S_n$ are the partial sums and $S = 1/(1-z)$ I tried to prove that $S_n - S \rightarrow 0$ by the following:

$|S_n - S|\leq \sum _{k=0}^{\infty}|z^k| + 1/|1-z|$ but here each term of the sum is $\leq 1$, but this doesn't help because I want to prove that this tends to zero.

Some help would be grateful.

Answer 1

The geometric series gives you: $$ S_k=\frac{1-z^{k+1}}{1-z} $$ for any $z\in\mathbb{C}\setminus\{1\}$. So $S_k-S =\frac{-z^{k+1}}{1-z}$ and $|S_k-S|=\frac{|z|^{k+1}}{|1-z|} \to 0$.

Answer 2

Nuking-mosquito solution: the function $f(z) = \frac1{1 - z}$ is holomorphic in $\{z\in\Bbb C:|z| < 1\}$, so analytic. The coefficients of the series around zero will be $f^{(n)}(0)/n!$, and you can easily calculate $$f^{(n)}(z) = \cdots$$ $$f^{(n)}(0) = \cdots$$