# 1/2 + 1/4 + 1/8 + 1/16 + ⋯

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

The sum of this series can be denoted in summation notation as:

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

As with any infinite series, the infinite sum

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

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

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

as n approaches infinity.

Multiplying sn by 2 reveals a useful relationship:

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

Subtracting sn from both sides,

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

As n approaches infinity, sn tends to 1.