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, every geometric 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

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

The sum of the first n terms of a geometric sequence is called the n-th partial sum (of the series); its formula is given below (S_n).

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.

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 more advanced mathematical texts "geometrical series" usually refers to the infinite series.

Examples

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

The partial sum S₅ follows thus (see the formula derived below)

$S_{5}\equiv 6+2+{\frac {2}{3}}+{\frac {2}{9}}+{\frac {2}{27}}=6\left[1+{\frac {1}{3}}+{\Big (}{\frac {1}{3}}{\Big )}^{2}+{\Big (}{\frac {1}{3}}{\Big )}^{3}+{\Big (}{\frac {1}{3}}{\Big )}^{4}\right]=6\left[{\frac {1-({\frac {1}{3}})^{5}}{1-{\frac {1}{3}}}}\right]={\frac {242}{27}}$

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

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

Summary: Convergence behaviour of the geometric series

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.

• For real q:

For q ≥ 1 the limit is +∞ or −∞ depending on the sign of a.

For q = −1 the series alternates between a and 0.

For q < −1 the sign of partial sums alternates, the limit of the absolute values of |S_n| is ∞, but no infinite limit exists.

• For complex q:

For |q| = 1 and q ≠ 1 (i.e., q = −1 or non-real complex) the partial sums S_n are bounded but not convergent.

For |q| > 1 and q non-real complex the partial sums S_n oscillate, the limit of the absolute values of |S_n| is ∞, but no infinite limit exists.