$\epsilon $ $\delta$ proof of geometric series $x$<1 Give an $\epsilon -\delta $ proof of  when $|x|<1$
 $$ \sum ^ \infty _{k=0} x^k= \frac{1}{1-x}  $$

Its been a more than a year from my analysis class.  Trying to recall def of convergence of a  partial sum that is 
$$\lim _{n \to \infty } S_n =s \equiv \forall \epsilon >0 , \exists N_\epsilon \in \mathbb{N}: \text{ if } n\geq N_\epsilon \Rightarrow |S_n-s|< \epsilon$$
not sure why the $\delta$ unless my memory has change it. 

Here is what I am thinking
setting $S_n=\sum^{n}_{k=0} x^k$ we gave 
$$\begin{aligned}
\left| \sum ^n _{k=0} x^k- \frac{1}{1-x} \right|
                   &=\left|(x^0+\dots x^n)- \frac{1}{1-x}\right|
                 \\  &=\left|\frac{1-x^{n+1}}{1-x}- \frac{1}{1-x}\right|
                 \\ &= \left|\frac{1}{1-x}-\frac{x^{n+1}}{1-x}-\frac{1}{1-x}\right|
\\&=\left|\frac{-x^{n+1}}{{1-x}}\right|
\end{aligned}$$
this is like cheating but  $\lim_{n \to \infty} (x^n)=0$. I need to use the Euclidean( or some greek i think i mean archimidean) prop of natural numbers for me to happy.

 A: As Arthur said in a comment $\varepsilon-\delta$ arguments is for continuity.
You're intuition is right. You were almost there.
$$\begin{aligned}
\lim\limits_{n \rightarrow \infty}\left| \sum ^n _{k=0} x^k- \frac{1}{1-x} \right|
                   &= \lim\limits_{n \rightarrow \infty}\left|(x^0+\dots x^n)- \frac{1}{1-x}\right|
                 \\  &= \lim\limits_{n \rightarrow \infty}\left|\frac{1-x^{n+1}}{1-x}- \frac{1}{1-x}\right|
                 \\ &= \lim\limits_{n \rightarrow \infty}\left|\frac{1}{1-x}-\frac{x^{n+1}}{1-x}-\frac{1}{1-x}\right|
\\&= \lim\limits_{n \rightarrow \infty}\left|\frac{-x^{n+1}}{{1-x}}\right|\\&= \lim\limits_{n \rightarrow \infty}\left|\frac{x^{n+1}}{{1-x}}\right|  \\&= \lim\limits_{n \rightarrow \infty}\frac{|x^{n+1}|}{|{1-x}|}  \\&= \lim\limits_{n \rightarrow \infty}\frac{|x|^{n+1}}{|{1-x}|} \\&= \frac{\lim\limits_{n \rightarrow \infty}|x|^{n+1}}{|{1-x}|} 
\end{aligned}$$
Finaly $\lim\limits_{n \rightarrow \infty}|x|^{n+1} = 0$ since $|x| < 1$.
A: If you want a formal proof that uses relatively little machinery, you can do this. 
We want to prove that if $|x|\lt 1$ then $\lim_{n\to\infty} |x|^n=0$.  This is obvious if $x=0$. So suppose that $x\ne 0$. Then $|x|=\frac{1}{1+t}$ for some positive $t$. 
Now using the Binomial Theorem, or the Bernoulli Inequality (easily proved by induction) we have $(1+t)^n \ge 1+nt\gt nt$. It follows that $|x|^n \lt \frac{1}{nt}$. To make $|x|^n \lt \epsilon$, it is sufficient to make $n\gt \frac{t}{\epsilon}$.
By the Archimedean property of the reals, there is an integer $N\gt \frac{t}{\epsilon}$. Then if $n\ge N$ we have $|x|^n\lt \epsilon$.
A: I just did this problem from a textbook which specifically asked for an $\epsilon-N$ proof, and then I came here to see how other people did it. Could I get some feedback on my solution?
Let $S_n$ be the finite sum
$$S_n=\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x}.$$
Let $\epsilon>0$. If $\epsilon>1/(1-x)$ then we can choose $N=1$, so $n\geq N$ implies
$$\left|S_n-\frac{1}{1-x}\right|=\frac{|x|^{n+1}}{1-x}\leq \frac{1}{1-x}=\epsilon.$$
For the remaining cases, let $\epsilon<1/(1-x)$, and set
$$N=\left\lceil{\frac{\log (\epsilon(1-x))}{\log |x|}}\right\rceil,$$
where the logarithms are both negative by construction, so $N$ is positive. Therefore, for all $n\geq N$, we have
$$\left|S_n-\frac{1}{1-x}\right|=\frac{|x|^{n+1}}{1-x}<\frac{|x|^\frac{\log (\epsilon(1-x))}{\log |x|}}{1-x}=\frac{\epsilon(1-x)}{1-x}=\epsilon.$$
