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.

In finance, since compound interest generates a geometric sequence, regular payments together with compound interest lead to a geometric 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 sum of the first 5 terms — the partial sum S₅ (see the formula derived below) — is for q = 1/3

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

and for q = −1/3

$S_{5}=6-2+{\frac {2}{3}}-{\frac {2}{9}}+{\frac {2}{27}}=6\left[1-{\frac {1}{3}}+{\Big (}{\frac {1}{3}}{\Big )}^{2}-{\Big (}{\frac {1}{3}}{\Big )}^{3}+{\Big (}{\frac {1}{3}}{\Big )}^{4}\right]=6\left[{\frac {1+({\frac {1}{3}})^{5}}{1+{\frac {1}{3}}}}\right]={\frac {122}{27}}$

Application in finance

When regular payments are combined with compound interest this generates a geometric series:

Regular deposits

If, for n time periods, a sum P is deposited at an interest rate of p percent, then — after the n-th period —

the first payment has increased to $P_{n}=P\left(1+{p \over 100}\right)^{n}$

the second to $P_{n-1}=P\left(1+{p \over 100}\right)^{n-1}$

etc., and the last one $P_{1}=P\left(1+{p \over 100}\right)$

Thus the cumulated sum

$P_{1}+P_{2}+\cdots P_{n}=Pq+Pq^{2}+\cdots +Pq^{n}\qquad {\text{where }}q=1+{p \over 100}$

is the n-th partial sum of a geometric series.

Regular down payments

If a loan L is to be payed off by n regular payments P, the total payment nP has to cover both the loan L and the accumulated interest I.

The interest for the payment at the end of the first time period is $I_{1}=P\left({p \over 100}\right)$,

for the payment after two time periods it is $I_{2}=P\left({p \over 100}\right)^{2}$,

etc., and for the last payment after n time periods the interest is $I_{n}=P\left({p \over 100}\right)^{n}$.

Thus the accumulated interest

$nP-L=I_{1}+I_{2}+\cdots +I_{n}=Pq+Pq^{2}+\cdots +Pq^{n}\qquad {\text{where }}q=1+{p \over 100}$

is the n-th partial sum of a geometric series. (From this equation, P can easily be calculated.)

Mathematical treatment

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

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

Partial sums

The partial sums of the series Σq^k are

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

because

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

Thus

$S_{n}=\sum _{k=1}^{n}a_{k}=a{\frac {1-q^{n}}{1-q}}{\text{ for }}q\neq 1{\text{ and }}S_{n}=an{\text{ for }}q=1$

Limit

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

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

Thus the sum or limit of the series is

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

Geometric power series

For each q, the geometric series is a series of numbers, but since — apart from the constant factor a — they all have the same form Σq^k, it is convenient to replace the quotient q by a variable x and consider the (real or complex) geometric power series (a series of functions):

$\sum _{k=1}^{\infty }x^{k}\ {\text{ for }}\ x\in \mathbb {R} \ {\text{ or }}\ \mathbb {C}$

The convergence radius of this power series is 1. It

• converges (more precisely: converges absolutely) for |x|<1 with the sum

$\sum _{k=1}^{\infty }x^{k}={1 \over 1-x}$

• and diverges for |x| ≥ 1.

• For real x:

For x ≥ 1 the limit is +∞.

For x = −1 the series alternates between 1 and 0.

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

• For complex x:

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

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