# Science:Infinite Series Module/Units/Unit 1/1.1 Infinite Sequences/1.1.11 Example rn

## Example

Determine the values of r so that the sequence

${\displaystyle \{r^{n}\}_{n=1}^{\infty }}$

is convergent.

## Complete Solution

We can solve this problem by considering cases for the value of r.

### Case 1

If r > 1, then  rn tends to infinity as n tends to infinity. The sequence is divergent in this case.

### Case 2

If r= 1, then

${\displaystyle \lim _{n\rightarrow a}r^{n}=1}$

so the sequence is convergent for this case.

### Case 3

If ${\displaystyle -1, then

${\displaystyle \lim _{n\rightarrow \infty }r^{n}=0,}$

so the sequence is convergent for this case.

### Case 4

If ${\displaystyle r=-1}$, then

${\displaystyle \lim _{n\rightarrow \infty }(-1)^{n}}$

does not exist, so the sequence is divergent for this case.

### Case 5

If ${\displaystyle r<-1}$, then ${\displaystyle r^{n}}$ tends to negative infinity as ${\displaystyle n}$ does not tend to a single finite number. The sequence is divergent in this case.

### Summary

Therefore, the sequence ${\displaystyle \{r^{n}\}_{n=1}^{\infty }}$ is convergent when ${\displaystyle -1.

## Explanation of Each Step

### Case 1

Consider the case when ${\displaystyle r=2}$. Then our sequence becomes

${\displaystyle \{2,4,8,16,32,64,\ldots \}}$

which tends to infinity.

### Case 2

Here we are using a fundamental property of limits, that the limit of a constant equals that constant:

${\displaystyle \lim _{x\rightarrow \infty }c=c}$

for any constant ${\displaystyle c}$ and ${\displaystyle x\in \mathbb {R} }$.

### Case 3

Consider the case when ${\displaystyle r=1/2}$. Then our sequence becomes

${\displaystyle \{1/2,1/4,1/8,1/16,1/32,\ldots \}}$

which tends to zero.

Similarly, if  ${\displaystyle r=-1/2}$. Then our sequence becomes

${\displaystyle \{-1/2,+1/4,-1/8,+1/16,-1/32,\ldots \}}$

which also tends to zero.

### Case 4

In this case, we have the sequence

${\displaystyle -1,+1,-1,+1,\ldots }$

As ${\displaystyle n}$ approaches infinity the sequence does not approach a unique value, so the limit does not exist.

### Case 5

This case is similar to Case 1. Consider the case when ${\displaystyle r=-2}$. Then our sequence becomes

${\displaystyle \{-2,+4,-8,+16,-32,+64,\ldots \}}$

The terms alternate between positive and negative numbers, and do not tend to a single finite number.

## Possible Challenge Areas

### Connecting Results to Definition of Convergence

In each of the cases, we used a limit to determine whether the sequence is convergent. According to our definition of convergence of a sequence, as long as our respective limits exist, then the sequence converges.