Symbolically:

${\displaystyle S=\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+...}$

Clearly,

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{n}}}=\sum _{n=1}^{\infty }{\bigg (}{\frac {1}{2}}{\bigg )}^{n}}$

This is a geometric series of the form ${\displaystyle \sum _{n=1}^{\infty }ar^{n-1}}$ where ${\displaystyle a={\frac {1}{2}}}$, ${\displaystyle r={\frac {1}{2}}}$, and ${\displaystyle S={\frac {a}{1-r}}}$. Therefore,

${\displaystyle S={\frac {\frac {1}{2}}{1-{\frac {1}{2}}}}={\frac {\frac {1}{2}}{\frac {1}{2}}}=1}$

...

${\displaystyle \sum _{n=\infty }^{1}{\frac {1}{2^{n}}}=\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}}$

In other words:

${\displaystyle ...+{\frac {1}{16}}+{\frac {1}{8}}+{\frac {1}{4}}+{\frac {1}{2}}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+...}$