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:

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