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Difference between revisions of "Geometric sequence"

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(→‎Proof)
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Let the sequence be  
 
Let the sequence be  
  
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<center><math>S=a_1+a_1r+a_1r^2+a_1r^3+\cdots</math></center>     
+
<center><math>S=a_1+a_1r+a_1r^2+a_1r^3+\cdots.</math></center>     
  
 
Multiplying by <math>r</math> yields,  
 
Multiplying by <math>r</math> yields,  
  
−
<center><math>S \cdot r=a_1r+a_1r^2+a_1r^3+\cdots</math></center>  
+
<center><math>S \cdot r=a_1r+a_1r^2+a_1r^3+\cdots.</math></center>  
  
 
We subtract these two equations to obtain:  
 
We subtract these two equations to obtain:  
  
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<center><math> S-Sr=a_1</math></center>  
+
<center><math> S-Sr=a_1.</math></center>  
  
 
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  
 
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  
Line 49: Line 49:
 
thus,  
 
thus,  
  
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<center><math>S=\frac{a_1}{1-r}</math></center>  
+
<center><math>S=\frac{a_1}{1-r}.</math></center>  
  
 
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).  
 
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).  

Revision as of 10:48, 19 December 2018

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:

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

with a fixed first term $a_1$ and common ratio $r$. Using this definition, the $n$th term has the closed-form:

$a_n = a_1\cdot r^{n-1}$

Summing a Geometric Sequence

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

$S_n = a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1}$

where $a_1$ is the first term in the sequence, and $r$ is the common ratio.

Proof

The geometric sequence can be rewritten as $a_1+r \cdot a_1+r^2 \cdot a_1+ \cdots + r^{n-1} \cdot a_1=a_1(1+r+r^2+ \cdots +r^{n-1})$ where $n$ is the amount of terms, $r$ is the common ratio, and $a_1$ is the first term. Multiplying in $(r-1)$ yields $r^n-1$ so $a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1}$.

Infinite Geometric Sequences

An infinite geometric sequence is a geometric sequence with an infinite number of terms. If the common ratio is small, the terms will approach 0 and 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:

$S = \frac{a_1}{1-r}$.



Where $a_1$ is the first term in the sequence, and $r$ is the common ratio.

Proof

Let the sequence be

$S=a_1+a_1r+a_1r^2+a_1r^3+\cdots.$

Multiplying by $r$ yields,

$S \cdot r=a_1r+a_1r^2+a_1r^3+\cdots.$

We subtract these two equations to obtain:

$S-Sr=a_1.$

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

$S(1-r)=a_1$

thus,

$S=\frac{a_1}{1-r}.$

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.

Problems

Intermediate

See also

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