I do not see how $ \frac { 1 }{ 1-x } = \sum _{ n=0 }^{ \infty }{ { x }^{ n } } $ I do not see how $$ \frac { 1 }{ 1-x } =\quad \sum _{ n=0 }^{ \infty  }{ { x }^{ n } }  $$

for example, When x = 10
and n = 1

Then $$ \frac { 1 }{ 1-10 } \neq \quad { 10 }^{ 1 } $$

What am overlooking? What is clear way to show? $$ \frac { 1 }{ 1-x } =\quad \sum _{ n=0 }^{ \infty  }{ { x }^{ n } }$$
First of all, you can't choose just one $n$ to plug into the formula. $n$ isn't a variable, it's an index: you sum over all $n$, meaning
$$
\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \dots,
$$
when the right hand side converges (meaning you can only plug in certain $x$). If $\left|x\right| &lt; 1$, you can show that this is true without much trouble: let $S_N = \sum_{n = 0}^N x^n$. Then $x S_N = \sum_{n = 0}^N x^{n+1}$. Then $S_N - x S_N = \sum_{n = 0}^N x^n - \sum_{n = 0}^N x^{n+1} = 1 - x^{N+1}$, and hence
$$
S_N = \frac{1 - x^{N+1}}{1 - x}.
$$
Taking the limit, we have
$$
\lim_{N\to\infty} S_N = \lim_{N\to\infty}\frac{1 - x^{N+1}}{1 - x} = \frac{1}{1 - x},
$$
because $x^N\to 0$ as $N\to\infty$ if $\left|x\right| &lt; 1$. By definition, $\sum_{n = 0}^{\infty} x^n = \lim_{N\to\infty} S_N$, so you have your formula.
$\sum _{n=0}^\infty x^n=1+x+x^2+\ldots $  which is a G.P. series with common ratio $x$ 

Hence $1+x+x^2+\ldots =\dfrac{1}{1-x} \text{if |x|&lt;1}$