Geometric series

A geometric series is a series associated with an infinite geometric sequence, i.e., the quotient q of two consecutive terms is the same for each pair.

A 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.

Example

The series

$6+2+{\frac {2}{3}}+{\frac {2}{9}}+{\frac {2}{27}}+\cdots$

is a geometric series with quotient

$q={\frac {1}{3}}$

and first term

$a=6$

and therefore its sum is

${6 \over 1-{\frac {1}{3}}}=9$

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.)