prove or disprove uniform convergence when $|x|<1$ $ \sum ^ \infty _{k=0} x^k= \frac{1}{1-x} $ prove or disprove uniform convergence 
when $|x|<1$
 $$ \sum ^ \infty _{k=0} x^k= \frac{1}{1-x}  $$

Def of uniform convergence of partial sums $\lim _{n \to \infty } S_n(x)=s$
$$\lim _{n \to \infty } S_n(x) =s \equiv \forall \epsilon >0 , \exists N_\epsilon \in \mathbb{N}: \text{ if } n\geq N_\epsilon \Rightarrow |S_n-s|< \epsilon \wedge \forall x\in (-1,1)$$
negation of uniform convergence being
$$ \exists \epsilon >0 , \forall N_\epsilon \in \mathbb{N}: \text{ if } n\geq N_\epsilon \Rightarrow |S_n-s|\geq  \epsilon \wedge \exists x\in (-1,1)$$

the bigest lead I have is that 
$$\lim_{x \to 1 }\frac{x^n}{1-x}=\frac{\lim x^n}{\lim (1-x)}=\frac{1}{0}=\infty $$
From previous question had $$\begin{aligned}
\left| \sum ^n _{k=0} x^k- \frac{1}{1-x} \right|
                   &=\left|(x^0+\dots+ x^n)- \frac{1}{1-x}\right|
                 \\  &=\left|\frac{1-x^{n+1}}{1-x}- \frac{1}{1-x}\right|
                 \\ &= \left|\frac{1}{1-x}-\frac{x^{n+1}}{1-x}-\frac{1}{1-x}\right|
\\&=\left|\frac{-x^{n+1}}{{1-x}}\right|
\end{aligned}$$
let $\epsilon = 1/2 $ and $x=.99999999$ so 
$$\left|\frac{-x^{n+1}}{{1-x}}\right|= .99999^{N+1}/{.000001}  $$
idk. Ill go over examples form my first real analysis class may be ill come back to me how to argue my point better
 A: $\frac{1}{1-x}$ is unbounded on $(-1,1)$, so the series does not converge uniformly on $(-1,1)$. Indeed we have the following theorem.

If $f_n:X \to \mathbb{R}$ is bounded on $X$ for every $n$ and $f_n$ converges uniformly to $f$ on $X$, then $f:X\to \mathbb{R}$ is bounded on $X$.

If the series converged uniformly, $\frac{1}{1-x}$ would have been bounded, so ad absurdum, it does not converge uniformly.

But it converges uniformly on $[-a,a]$ for every $0\le a <1$. For $|x| \le a$, we have :
$$|S_n(x)-\frac{1}{1-x}|=\frac{|x|^{n+1}}{1-x}\le \frac{|a|^{n+1}}{1-a}$$
The last term tends to $0$. it follows that $|\sum ^ n _{k=0} x^k - \frac{1}{1-x}|< \epsilon$ for all $x \in [-a,a]$ and all $n$ big enough (I will let you do eplain in more details this part).
Finally you have uniform convergence of the series on $[-a,a]$.
A: You're very close. Let $S_n(x)$ denote the partial sum $\sum_{k=0}^n x_k$ and let $S_\infty(x)$ be the pointwise limit $\frac{1}{1-x}$. Now you showed that 
$$\tag{$*$} |S_n(x) - S_\infty (x) |=\frac{x^{n+1}}{1-x}.$$
To disprove uniform convergence over the open interval $(-1,1)$, we will  show that there exists $\epsilon>0$ such that for every $n$, there exists $x_n\in (-1,1)$ such that $|S_n (x_n) - S_\infty(x_n)|\ge \epsilon$ for infinitely many $n$'s. Choose $x_n =1-1/(n+1)$.  Then the RHS of $(*)$ is 
$$ (n+1)\times (1-\frac{1}{1+n})^{n+1}.$$
However $(1- \frac{1}{n+1})^{n+1}\to e^{-1}$. Therefore, 
$$|S_n(x_n)-S_n(x_n)|\to \infty.$$
Note The correct negation of uniform convergence is the following. 
There exists $\epsilon>0$ such that for every $N\in{\mathbb N}$ there exists $n\ge N$ and $x\in (-1,1)$ such that $|S_n(x) - S_{\infty}(x)|\ge \epsilon$. 
(difference from what was listed in the question: not for all $n$ large,  but for infinitely many $n$-s).   
