User:Yafujifide

Symbolically:

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

Clearly,

$\sum_{n=1}^\infty \frac{1}{2^{n}}=\sum_{n=1}^\infty \bigg(\frac{1}{2}\bigg)^{n}$

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

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

...

$\sum_{n=\infty}^1 \frac{1}{2^{n}}=\sum_{n=1}^\infty \frac{1}{2^{n}}$

In other words:

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