Well, we have that
$s_n=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$
Now, we look at $n\to \infty$
$\lim_{n\to\infty}\frac{1-x^{n+1}}{1-x}=\lim_{n\to\infty}\sum_{k=0}^{n}x^k$
For what $x$ does this limit exist?
Clearly, we're only concerned about $x^{n+1}$. If $|x|>1$, then $x^{n+1}$ grows large, and there is no limit. If $|x|<1$, then $x-1\neq 0$ and $x^{n+1}$ goes to $0$ as $n\to\infty$, so in that case, we can assert that
$\lim_{n\to\infty}\frac{1-x^{n+1}}{1-x}=\frac 1{1-x}=\sum_{k=0}^{\infty}x^k$
If $x=1$, $x-1=0$ and we find ourselves in trouble. However, we can say that $\sum\limits_{k = 0}^n {{1^k}} = n$ in which case the sequence of partial sums has no limits. Finally, for $x=-1$, we have $(-1)^{n+1}$ which oscillates and has no limit.
Thus, $\frac{1}{{1 - x}} = \sum\limits_{k = 0}^\infty {{x^k}} $ makes sense only for $|x|<1$.