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. Thus, the series has the form

$a+aq+aq^{2}+aq^{3}+\cdots$

where the quotient (ratio) of the (n+1)th and the nth term is

${\frac {aq^{n}}{aq^{n-1}}}=q.$

An infinite geometric series (i.e., a series with an infinite number of terms) converges if and only if |q|<1, in which case its sum is $a \over 1-q$, where a is the first term of the series.

Remarks

1. The sum of a finite (n) terms of a geometric sequence is a finite number S_n; its formula is given below.
2. 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 more advanced 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

By definition, a geometric series

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

can be written as

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

where

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

The partial sums of the power series Σq^k are

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

because

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

Since

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

it is

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

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

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

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