Geometric series

A geometric series is a series associated with a geometric sequence, i.e., the ratio (or quotient) q of two consecutive terms is the same for each pair.

An infinite geometric series converges if and only if |q|<1.

Then its sum is $a \over 1-q$ where a is the first term of the series.

Remark
Since every finite geometric sequence is the initial segment of a uniquely determined infinite geometric sequence every finite geometric series is the initial segment of a corresponding infinite geometric series. Therefore, while in elementary mathematics the difference between "finite" and "infinite" may be stressed, in mathematical texts "geometrical series" usually refers to the infinite series.

Examples

Positive ratio		Negative ratio
The series $6+2+{\frac {2}{3}}+{\frac {2}{9}}+{\frac {2}{27}}+\cdots$ and corresponding sequence of partial sums $6,8,{\frac {26}{3}},{\frac {80}{9}},{\frac {242}{27}},\cdots$ is a geometric series with quotient $q={\frac {1}{3}}$ and first term $a=6$ and therefore its sum is ${6 \over 1-{\frac {1}{3}}}=9$		The series $6-2+{\frac {2}{3}}-{\frac {2}{9}}+{\frac {2}{27}}-+\cdots$ and corresponding sequence of partial sums $6,4,{\frac {14}{3}},{\frac {40}{9}},{\frac {122}{27}},\cdots$ is a geometric series with quotient $q=-{\frac {1}{3}}$ and first term $a=6$ and therefore its sum is ${6 \over 1-\left(-{\frac {1}{3}}\right)}={\frac {9}{2}}$

Power series

Any geometric series

$\sum _{k=1}^{\infty }a_{k}\qquad (a_{k}\in \mathbb {C} )$

can be written as

$a\sum _{k=0}^{\infty }x^{k}$

where

$a=a_{1}\qquad {\textrm {and}}\qquad x={a_{k+1} \over a_{k}}\in \mathbb {C} {\hbox{ is the constant quotient}}$

The partial sums of the power series Σx^k are

$S_{n}=\sum _{k=0}^{n-1}x^{k}=1+x+x^{2}+\cdots +x^{n-1}={\begin{cases}{\displaystyle {\frac {1-x^{n}}{1-x}}}&{\hbox{for }}x\neq 1\\n\cdot 1&{\hbox{for }}x=1\end{cases}}$

because

$(1-x)(1+x+x^{2}+\cdots +x^{n-1})=1-x^{n}$

Since

$\lim _{n\to \infty }{1-x^{n} \over 1-x}={1-\lim _{n\to \infty }x^{n} \over 1-x}\quad (x\neq 1)$

it is

$\lim _{n\to \infty }S_{n}={1 \over 1-x}\quad \Leftrightarrow \quad |x|<1$

and the geometric series converges (more precisely: converges absolutely) for |x|<1 with the sum

$\sum _{k=1}^{\infty }a_{k}={a \over 1-q}$

and diverges for |x| ≥ 1. (Depending on the sign of a, the limit is +∞ or −∞ for x≥1.)