Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Geometric sequence"
m (Reverted edits by IntrepidMath (IntrepidMath); changed back to last version by JBL)
Line 51: Line 51:
 
One common instance of summing infinite geometric sequences is the [[decimal expansion]] of most [[rational number]]s.  For instance, <math>0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots</math> has first term <math>a_0 = \frac 3{10}</math> and common ratio <math>\frac1{10}</math>, so the infinite sum has value <math>S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13</math>, just as we would have expected.
 
One common instance of summing infinite geometric sequences is the [[decimal expansion]] of most [[rational number]]s.  For instance, <math>0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots</math> has first term <math>a_0 = \frac 3{10}</math> and common ratio <math>\frac1{10}</math>, so the infinite sum has value <math>S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13</math>, just as we would have expected.
  
−
A somewhat interesting application is applying this formula to the series 1, -1, 1...  We get: <math>\displaystyle \frac{1}{1-(-1)}=\frac{1}{2}</math>.  This is rather odd because the partial sums never equal to 1/2.
 
 
==See Also==
 
==See Also==
 
*[[arithmetic sequence|Arithmetic Sequences]]
 
*[[arithmetic sequence|Arithmetic Sequences]]

Revision as of 09:25, 7 July 2006

Definition

A geometric sequence is a sequence of numbers in which each term is a fixed multiple of the previous term. For example: 1, 2, 4, 8, 16, 32, ... is a geometric sequence because each term is twice the previous term. In this case, 2 is called the common ratio of the sequence. More formally, a geometric sequence may be defined recursively by:

$a_n = r\cdot a_{n-1}, n \geq 1$

with a fixed $a_0$ and common ratio $r$. Using this definition, the $n$th term has the closed-form:

$\displaystyle a_n = a_0\cdot r^n$

Summing a Geometric Sequence

The sum of the first $n$ terms of a geometric sequence is given by

$S_n = a_0 + a_1 + \ldots + a_{n - 1} = a_0\cdot\frac{r^n-1}{r-1}$

where $a_0$ is the first term in the sequence, and $r$ is the common ratio.

Infinite Geometric Sequences

An infinite geometric sequence is a geometric sequence with an infinite number of terms. If the common ratio is small, the sum of the terms will approach a fixed limit. In this case, "small" means $|r|<1$. We say that the sum of the terms of this sequence is a convergent sum.

For instance, the series $1 + \frac12 + \frac14 + \frac18 + \cdots$, sums to 2. The general formula for the sum of such a sequence is:

$S = \frac{a_0}{1-r}$.



Where $a_0$ is the first term in the sequence, and $r$ is the common ratio.

"Proof": Let the sequence be

$S=a_0+a_0r+a_0r^2+a_0r^3+\cdots$

Multiplying by $r$ yields,

$S \cdot r=a_0r+a_0r^2+a_0r^3+\cdots$

We subtract these two equations to obtain:

$S-S\cdot r=a_0$

There is only one term on the RHS because the rest of the terms cancel out after subtraction. Finally, we can factor and divide to get

$\displaystyle S(1-r)=a_0$

thus,

$S=\frac{a_0}{1-r}$

This method of multiplying the sequence and subtracting equations, called telescoping, is a frequently used method to evaluate infinite sequences. In fact, the same method can be used to calculate the sum of a finite geometric sequence (given above).

One common instance of summing infinite geometric sequences is the decimal expansion of most rational numbers. For instance, $0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots$ has first term $a_0 = \frac 3{10}$ and common ratio $\frac1{10}$, so the infinite sum has value $S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13$, just as we would have expected.

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Geometric_sequence&oldid=6220"