# 1/2 + 1/4 + 1/8 + 1/16 + ⋯

First six summands drawn as portions of a square.
The geometric series on the real line.

In mathematics, the infinite series 1/2 + 1/4 + 1/8 + 1/16 + · · · is an elementary example of a geometric series that converges absolutely.

There are many expressions that can be shown to be equivalent to the problem, such as the form: 2−1 + 2−2 + 2−3...

Its sum is

${\displaystyle {\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots =\sum _{n=1}^{\infty }\left({\frac {1}{2}}\right)^{n}={\frac {\frac {1}{2}}{1-{\frac {1}{2}}}}=1.}$

## Proof

As with any infinite series, the infinite sum

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

is defined to mean the limit of the sum of the first n terms

${\displaystyle s_{n}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots +{\frac {1}{2^{n-1}}}+{\frac {1}{2^{n}}}}$

as n approaches infinity.

Multiplying sn by 2 reveals a useful relationship:

${\displaystyle 2s_{n}={\frac {2}{2}}+{\frac {2}{4}}+{\frac {2}{8}}+{\frac {2}{16}}+\cdots +{\frac {2}{2^{n}}}=1+\left[{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+\cdots +{\frac {1}{2^{n-1}}}\right]=1+\left[s_{n}-{\frac {1}{2^{n}}}\right].}$

Subtracting sn from both sides,

${\displaystyle s_{n}=1-{\frac {1}{2^{n}}}.}$

As n approaches infinity, sn tends to 1.