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 $\displaystyle \left | \frac{z^{n+1}}{1-z} \right | \le \frac{(1-\delta)^n}{\delta} = \epsilon$
 A: 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 $\frac{1}{1-z}$.
A: Claim: If $p(x) = \sum_{n=0}^{\infty} c_n(x-a)^n$ is a real power series that converges for some $x_1\neq a$, then the series converges uniformly on $\left| x - a\right|\leq r$, for all $r < \left| x_1 - a\right|$.
Proof: We know that $\displaystyle\sum_n c_n (x_1 - a)^n$ converges, so $\left| c_n\right|\left|x_1 - a\right|^n\leq B$ for all $n$ (some $B\in\Bbb{R}^+$). Let $u_n(x) = c_n(x - a)^n$ and $M_n = B\left(r/\left|x_1 - a\right|\right)^n$. I claim that we can apply the $M$-test. Note that $\left|u_n(x)\right|\leq M_n$ for all $x$ such that $\left| x - a\right|\leq r$. Now
\begin{align*}
\left|u_n(x)\right| &= \left|c_n\right|\left|x - a\right|^n\\
&\leq \left|c_n\right|r^n\\
&= \left|c_n\right| M_n\frac{\left|x_1 - a\right|^n}{B}\\
&\leq M_n\quad\checkmark
\end{align*}
(as $B\geq\left|c_n\right|\left|x_1 - a\right|^n$). Then we have $M_n = B\rho^n$, where $\rho = \frac{r}{\left|x_1 - a\right|}$, and $\left|\rho\right| < 1$, so that $\displaystyle\sum_n M_n < \infty$. By the Weierstrass $M$-test, the convergence of our original series is uniform, and we are done.
This implies that your series converges uniformly on any real interval ball $\left|x - a\right|\leq r$, where $r < 1$. One would probably be able to modify the proof to apply to complex power series as well, to show that your series converges. Edit: In fact, the same proof works for complex power series as well.
A: You got
$$\left|s_n(z)-{1\over 1-z}\right|=\left|{z^{n+1}\over1-z}\right|\qquad(n\geq0)\ .$$
For uniform convergence of your series you would need that the right side converges uniformly to $0$ on $D: \ |z|<1$. This is not the case even if you forget about the denominator becoming zero at $z=1$: One has
$$\left|{z^{n+1}\over1-z}\right|\geq{1\over2}|z|^{n+1}\ .$$
Now choose
$$z_n:=-2^{-1/(n+1)}\in D\qquad(n\geq1)\ .$$
Then
$$\left|{z_n^{n+1}\over1-z_n}\right|\geq{1\over4}\qquad(n\geq1)\ .$$
This shows that the condition of uniform convergence in $D$ cannot be fulfilled even for $\epsilon:={1\over4}$.
