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Is this sequence increasing or decreasing? Is it bounded? How do I more formally show the bound?

0

$begingroup$

So I have this general equation for a sequence:

$$a_n = frac1-n2+n$$

is it increasing or decraesing? Writing out the first few terms I have:

$$0, frac-14, frac-25,frac-36,frac-47,frac-58$$

using a derivative test, we get:

$$f(x)= frac1-x2+x$$

$$f'(x)= -1$$

so the sequence is decraesing as $n -> infty$

Since the first term is 0, it's bounded above by 0. And it looks like it never crosses -1 so it's bounded below by -1. But how do I show this more vigorously?

Taking the limits seems like half the answer. What if the function oscillates around the limit but cross it? Taking the limit doesn't seem like the complete way to find the bounds right?

sequences-and-series

share|cite|improve this question

asked Mar 28 at 1:01

Jwan622Jwan622

2,33411632

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Show $a_n+1-a_n < 0$.
  $endgroup$
  – amsmath
  Mar 28 at 1:04
  

  
  
  
  

add a comment | 

0

$begingroup$

So I have this general equation for a sequence:

$$a_n = frac1-n2+n$$

is it increasing or decraesing? Writing out the first few terms I have:

$$0, frac-14, frac-25,frac-36,frac-47,frac-58$$

using a derivative test, we get:

$$f(x)= frac1-x2+x$$

$$f'(x)= -1$$

so the sequence is decraesing as $n -> infty$

Since the first term is 0, it's bounded above by 0. And it looks like it never crosses -1 so it's bounded below by -1. But how do I show this more vigorously?

Taking the limits seems like half the answer. What if the function oscillates around the limit but cross it? Taking the limit doesn't seem like the complete way to find the bounds right?

sequences-and-series

share|cite|improve this question

asked Mar 28 at 1:01

Jwan622Jwan622

2,33411632

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Show $a_n+1-a_n < 0$.
  $endgroup$
  – amsmath
  Mar 28 at 1:04
  

  
  
  
  

add a comment | 

$begingroup$

So I have this general equation for a sequence:

$$a_n = frac1-n2+n$$

is it increasing or decraesing? Writing out the first few terms I have:

$$0, frac-14, frac-25,frac-36,frac-47,frac-58$$

using a derivative test, we get:

$$f(x)= frac1-x2+x$$

$$f'(x)= -1$$

so the sequence is decraesing as $n -> infty$

Since the first term is 0, it's bounded above by 0. And it looks like it never crosses -1 so it's bounded below by -1. But how do I show this more vigorously?

Taking the limits seems like half the answer. What if the function oscillates around the limit but cross it? Taking the limit doesn't seem like the complete way to find the bounds right?

sequences-and-series

share|cite|improve this question

asked Mar 28 at 1:01

Jwan622Jwan622

2,33411632

$endgroup$

share|cite|improve this question

asked Mar 28 at 1:01

Jwan622Jwan622

2,33411632

share|cite|improve this question

asked Mar 28 at 1:01

Jwan622Jwan622

2,33411632

asked Mar 28 at 1:01

Jwan622Jwan622

2,33411632

asked Mar 28 at 1:01

Jwan622Jwan622

2,33411632

2 Answers
2

active

oldest

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2

$begingroup$

Hint: Write it as $a_n=-1+frac 32+n$ and it is clearer.

share|cite|improve this answer

answered Mar 28 at 1:10

Ross MillikanRoss Millikan

300k24200375

$endgroup$

add a comment | 

2

$begingroup$

Note that

$$a_n = frac1-n2+n = frac-2-n+32+n = -1 + frac32+n tag1labeleq1$$

Thus, for $n gt -2$, $a_n gt -1$. Also, this allows you to determine that $lim_n to inftya_n = -1$ since $frac32+n$ goes to $0$.

share|cite|improve this answer

answered Mar 28 at 1:10

John OmielanJohn Omielan

4,4512215

$endgroup$

add a comment | 

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2

$begingroup$

Hint: Write it as $a_n=-1+frac 32+n$ and it is clearer.

share|cite|improve this answer

answered Mar 28 at 1:10

Ross MillikanRoss Millikan

300k24200375

$endgroup$

add a comment | 

2

$begingroup$

Hint: Write it as $a_n=-1+frac 32+n$ and it is clearer.

share|cite|improve this answer

answered Mar 28 at 1:10

Ross MillikanRoss Millikan

300k24200375

$endgroup$

add a comment | 

$begingroup$

Hint: Write it as $a_n=-1+frac 32+n$ and it is clearer.

share|cite|improve this answer

answered Mar 28 at 1:10

Ross MillikanRoss Millikan

300k24200375

$endgroup$

share|cite|improve this answer

answered Mar 28 at 1:10

Ross MillikanRoss Millikan

300k24200375

answered Mar 28 at 1:10

Ross MillikanRoss Millikan

300k24200375

answered Mar 28 at 1:10

Ross MillikanRoss Millikan

300k24200375

2

$begingroup$

Note that

$$a_n = frac1-n2+n = frac-2-n+32+n = -1 + frac32+n tag1labeleq1$$

Thus, for $n gt -2$, $a_n gt -1$. Also, this allows you to determine that $lim_n to inftya_n = -1$ since $frac32+n$ goes to $0$.

share|cite|improve this answer

answered Mar 28 at 1:10

John OmielanJohn Omielan

4,4512215

$endgroup$

add a comment | 

2

$begingroup$

Note that

$$a_n = frac1-n2+n = frac-2-n+32+n = -1 + frac32+n tag1labeleq1$$

Thus, for $n gt -2$, $a_n gt -1$. Also, this allows you to determine that $lim_n to inftya_n = -1$ since $frac32+n$ goes to $0$.

share|cite|improve this answer

answered Mar 28 at 1:10

John OmielanJohn Omielan

4,4512215

$endgroup$

add a comment | 

$begingroup$

Note that

$$a_n = frac1-n2+n = frac-2-n+32+n = -1 + frac32+n tag1labeleq1$$

Thus, for $n gt -2$, $a_n gt -1$. Also, this allows you to determine that $lim_n to inftya_n = -1$ since $frac32+n$ goes to $0$.

share|cite|improve this answer

answered Mar 28 at 1:10

John OmielanJohn Omielan

4,4512215

$endgroup$

share|cite|improve this answer

answered Mar 28 at 1:10

John OmielanJohn Omielan

4,4512215

answered Mar 28 at 1:10

John OmielanJohn Omielan

4,4512215

answered Mar 28 at 1:10

John OmielanJohn Omielan

4,4512215