Symbolically:
S = ∑ n = 1 ∞ 1 2 n = 1 2 + 1 4 + 1 8 + 1 16 + . . . {\displaystyle S=\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+...}
Clearly,
∑ n = 1 ∞ 1 2 n = ∑ n = 1 ∞ ( 1 2 ) n {\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 ∑ n = 1 ∞ a r n − 1 {\displaystyle \sum _{n=1}^{\infty }ar^{n-1}} where a = 1 2 {\displaystyle a={\frac {1}{2}}} , r = 1 2 {\displaystyle r={\frac {1}{2}}} , and S = a 1 − r {\displaystyle S={\frac {a}{1-r}}} . Therefore,
S = 1 2 1 − 1 2 = 1 2 1 2 = 1 {\displaystyle S={\frac {\frac {1}{2}}{1-{\frac {1}{2}}}}={\frac {\frac {1}{2}}{\frac {1}{2}}}=1}
...
∑ n = ∞ 1 1 2 n = ∑ n = 1 ∞ 1 2 n {\displaystyle \sum _{n=\infty }^{1}{\frac {1}{2^{n}}}=\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}}
In other words:
. . . + 1 16 + 1 8 + 1 4 + 1 2 = 1 2 + 1 4 + 1 8 + 1 16 + . . . {\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}}+...}
where 
, 
, and 
. Therefore,
