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. The sum of the series is 1. In summation notation, this may be expressed as

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

First six summands drawn as portions of a square.

The geometric series on the real line.

The series is related to philosophical questions considered in antiquity, particularly to Zeno's paradoxes.

Proof

As with any infinite series, the sum

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

is defined to mean the limit of the partial 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. By various arguments,[lower-alpha 1] one can show that this finite sum is equal to

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

As $n$ approaches infinity, the term ${\frac {1}{2^{n}}}$ approaches 0 and so $s n$ tends to 1.

History

Zeno's paradox

This series was used as a representation of many of Zeno's paradoxes.[1] For example, in the paradox of Achilles and the Tortoise, the warrior Achilles was to race against a tortoise. The track is 100 meters long. Achilles could run at 10 m/s, while the tortoise only 5. The tortoise, with a 10-meter advantage, Zeno argued, would win. Achilles would have to move 10 meters to catch up to the tortoise, but by then, the tortoise would already have moved another five meters. Achilles would then have to move 5 meters, where the tortoise would move 2.5 meters, and so on. Zeno argued that the tortoise would always remain ahead of Achilles.

The Eye of Horus

The parts of the Eye of Horus were once thought to represent the first six summands of the series.[2]

In a myriad ages it will not be exhausted

A version of the series appears in the ancient Taoist book Zhuangzi. The miscellaneous chapters "All Under Heaven" include the following sentence: "Take a chi long stick and remove half every day, in a myriad ages it will not be exhausted."

References

Wachsmuth, Bet G. "Description of Zeno's paradoxes". Archived from the original on 2014-12-31. Retrieved 2014-12-29.
Stewart, Ian (2009). Professor Stewart's Hoard of Mathematical Treasures. Profile Books. pp. 76–80. ISBN 978 1 84668 292 6.

For example: multiplying $s n$ by 2 yields $2s_{n}={\frac {2}{2}}+{\frac {2}{4}}+{\frac {2}{8}}+{\frac {2}{16}}+\cdots +{\frac {2}{2^{n}}}=1+\left[{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+\cdots +{\frac {1}{2^{n-1}}}\right]=1+\left[s_{n}-{\frac {1}{2^{n}}}\right].$ Subtracting $s n$ from both sides, one concludes $s_{n}=1-{\frac {1}{2^{n}}}.$ Other arguments might proceed by mathematical induction, or by adding ${\frac {1}{2^{n}}}$ to both sides of $s_{n}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots +{\frac {1}{2^{n-1}}}+{\frac {1}{2^{n}}}$ and manipulating to show that the right side of the result is equal to 1.

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[2] Wachsmuth, Bet G. "Description of Zeno's paradoxes". Archived from the original on 2014-12-31. Retrieved 2014-12-29.

[3] Stewart, Ian (2009). Professor Stewart's Hoard of Mathematical Treasures. Profile Books. pp. 76–80. ISBN 978 1 84668 292 6.

[1] For example: multiplying $s n$ by 2 yields $2s_{n}={\frac {2}{2}}+{\frac {2}{4}}+{\frac {2}{8}}+{\frac {2}{16}}+\cdots +{\frac {2}{2^{n}}}=1+\left[{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+\cdots +{\frac {1}{2^{n-1}}}\right]=1+\left[s_{n}-{\frac {1}{2^{n}}}\right].$ Subtracting $s n$ from both sides, one concludes $s_{n}=1-{\frac {1}{2^{n}}}.$ Other arguments might proceed by mathematical induction, or by adding ${\frac {1}{2^{n}}}$ to both sides of $s_{n}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots +{\frac {1}{2^{n-1}}}+{\frac {1}{2^{n}}}$ and manipulating to show that the right side of the result is equal to 1.