<s>[INST] showing $1 + z + z^2 + \dots $ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$ What test can I use to show that $1 + z + z^2 + \dots $ uniformly converges to $\frac{1}{1-z}$ for $|z| < 1$.
I know $\displaystyle 1 + z + z^2 + \dots +z^n = \frac{1-z^{n+1}}{1-z}$ and as $n \to \infty$, $1 + z + z^2 + \dots = \frac{1}{1-z}$ for $|z| < 1$ but how to show uniform convergence using it's definition?
ADDED::
May be we could use something like this, not sure though ... please suggest correction.
$$\left |S_n(z) - \frac{1}{1-z} \right | = \left |\frac{1-z^{n+1}}{1-z} - \frac{1}{1-z} \right | =\left | \frac{z^{n+1}}{1-z} \right |$$
If $|1-z| = \delta : 0< \delta < 1 \implies 1-|z| \le \delta $ and we have $\displaystyle \left | \frac{z^{n+1}}{1-z} \right | \le \frac{(1-\delta)^n}{\delta} = \epsilon$
 [/INST] Let denote 
$$S_n(z)=\sum_{k=0}^n z^k=\frac{1-z^{n+1}}{1-z}$$
the partial sum of the series so we have
$$\left|R_n(z)\right|=\left|S_n(z)-\sum_{k=0}^\infty z^k\right|=\frac{|z|^{n+1}}{|1-z|}$$
so it's clear that
$$\sup_{|z|<1}|R_n(z)|=+\infty$$
so $(S_n(z))$ does not converge uniformly to $\frac{1}{1-z}$.
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