Geometric sequence

(Redirected from Geometric series)

In algebra, a geometric sequence, sometimes called a geometric progression, is a sequence of numbers such that the ratio between any two consecutive terms is constant. This constant is called the common ratio of the sequence.

For example, $1, 2, 4, 8$ is a geometric sequence with common ratio $2$ and $100, -50, 25, -25/2$ is a geometric sequence with common ratio $-1/2$; however, $1, 3, 9, -27$ and $-3, 1, 5, 9, \ldots$ are not geometric sequences, as the ratio between consecutive terms varies.

More formally, the sequence $a_1, a_2, \ldots , a_n$ is a geometric progression if and only if $a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}$. A similar definition holds for infinite geometric sequences. It appears most frequently in its three-term form: namely, that constants $a$, $b$, and $c$ are in geometric progression if and only if $b / a = c / b$.


Because each term is a common multiple of the one before it, every term of a geometric sequence can be expressed as the sum of the first term and a multiple of the common ratio. Let $a_1$ be the first term, $a_n$ be the $n$th term, and $r$ be the common ratio of any geometric sequence; then, $a_n = a_1 r^{n-1}$.

A common lemma is that a sequence is in geometric progression if and only if $a_n$ is the geometric mean of $a_{n-1}$ and $a_{n+1}$ for any consecutive terms $a_{n-1}, a_n, a_{n+1}$. In symbols, $a_n^2 = a_{n-1}a_{n+1}$. This is mostly used to perform substitutions, though it occasionally serves as a definition of geometric sequences.


A geometric series is the sum of all the terms of a geometric sequence. They come in two varieties, both of which have their own formulas: finitely or infinitely many terms.


A finite geometric series with first term $a_1$, common ratio $r$ not equal to one, and $n$ total terms has a value equal to $\frac{a_1(r^n-1)}{r-1}$.

Proof: Let the geometric series have value $S$. Then \[S = a_1 + a_1r + a_1r^2 + \cdots + a_1r^{n-1}.\] Factoring out $a_1$, mulltiplying both sides by $(r-1)$, and using the difference of powers factorization yields \[S(r-1) = a_1(r-1)(1 + r + r^2 + \cdots + r^{n-1}) = a_1(r^n-1).\] Dividing both sides by $r-1$ yields $S=\frac{a_1(r^n-1)}{r-1}$, as desired. $\square$


An infinite geometric series converges if and only if $|r|<1$; if this condition is satisfied, the series converges to $\frac{a_1}{1-r}$.

Proof: The proof that the series convergence if and only if $|r|<1$ is an easy application of the ratio test from calculus; thus, such a proof is beyond the scope of this article. If one assumes convergence, there is an elementary proof of the formula that uses telescoping. Using the terms defined above, \[S = a_1 + a_1r + a_1r^2 + \cdots.\] Multiplying both sides by $r$ and adding $a_1$, we find that \[rS + a_1 = a_1 + r(a_1 + a_1r + \cdots) = a_1 + a_1r + a_1r^2 + \cdots = S.\] Thus, $rS + a_1 = S$, and so $S = \frac{a_1}{1-r}$. $\square$


Here are some problems with solutions that utilize geometric sequences and series.


See also