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Sequence defined by $a_n+1 = sqrt2+a_n$, $a_1 = sqrt2$ is monotonically increasing

0

1

$begingroup$

I am trying to prove that the sequence $a_1=sqrt2$, $a_n+1=sqrt2+a_n$ is monotonically increasing.
My thought was that since $a_n+1^2= 2+a_n> 2a_n > a_n times a_n > a_n^2, a_n+1 > a_n^2$.

However, upon reflecting I think this is not actually true that just because $a_n+1^2 > a_n^2, a_n+1 > a_n$.

Is the only way to prove this by using induction? I was trying to do something more direct. If it is using induction, could you outline the proof?

sequences-and-series

share|cite|improve this question

edited Mar 26 at 6:52

Theo Bendit

21k12355

asked Mar 26 at 5:32

jacksonfjacksonf

14711

$endgroup$

add a comment | 

0

1

$begingroup$

I am trying to prove that the sequence $a_1=sqrt2$, $a_n+1=sqrt2+a_n$ is monotonically increasing.
My thought was that since $a_n+1^2= 2+a_n> 2a_n > a_n times a_n > a_n^2, a_n+1 > a_n^2$.

However, upon reflecting I think this is not actually true that just because $a_n+1^2 > a_n^2, a_n+1 > a_n$.

Is the only way to prove this by using induction? I was trying to do something more direct. If it is using induction, could you outline the proof?

sequences-and-series

share|cite|improve this question

edited Mar 26 at 6:52

Theo Bendit

21k12355

asked Mar 26 at 5:32

jacksonfjacksonf

14711

$endgroup$

add a comment | 

$begingroup$

I am trying to prove that the sequence $a_1=sqrt2$, $a_n+1=sqrt2+a_n$ is monotonically increasing.
My thought was that since $a_n+1^2= 2+a_n> 2a_n > a_n times a_n > a_n^2, a_n+1 > a_n^2$.

However, upon reflecting I think this is not actually true that just because $a_n+1^2 > a_n^2, a_n+1 > a_n$.

Is the only way to prove this by using induction? I was trying to do something more direct. If it is using induction, could you outline the proof?

sequences-and-series

share|cite|improve this question

edited Mar 26 at 6:52

Theo Bendit

21k12355

asked Mar 26 at 5:32

jacksonfjacksonf

14711

$endgroup$

share|cite|improve this question

edited Mar 26 at 6:52

Theo Bendit

21k12355

asked Mar 26 at 5:32

jacksonfjacksonf

14711

share|cite|improve this question

edited Mar 26 at 6:52

Theo Bendit

21k12355

asked Mar 26 at 5:32

jacksonfjacksonf

14711

edited Mar 26 at 6:52

Theo Bendit

21k12355

edited Mar 26 at 6:52

Theo Bendit

21k12355

asked Mar 26 at 5:32

jacksonfjacksonf

14711

asked Mar 26 at 5:32

jacksonfjacksonf

14711

2 Answers
2

active

oldest

votes

1

$begingroup$

Hint: All quantities involved are positive, and $a_n+1=sqrt2+a_n>a_n$ if and only if $a_n^2<a_n+2$, i.e. $(a_n-2)(a_n+1)<0$. Can you put a bound on $a_n$ and hence use this inequality to prove the result?

share|cite|improve this answer

edited Mar 26 at 5:55

answered Mar 26 at 5:49

YiFanYiFan

5,3832728

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, an edit was made to the post that was actually incorrect. It is increasing. √2 < √(2+√2) and so on
  $endgroup$
  – jacksonf
  Mar 26 at 5:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have already proved that an is always less than 2. I have to say I'm still confused, how does this bound help prove the result?
  $endgroup$
  – jacksonf
  Mar 26 at 5:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $(a_n-2)(a_n+1)<0$ if and only if $a<2$ and $a_n>-1$. You have the first part, the second part is trivial. So the inequality is true!
  $endgroup$
  – YiFan
  Mar 26 at 6:00
  

  
  
  
  

add a comment | 

1

$begingroup$

First you check as an exercise that $a_n>0$ for all $ngeq 1$

We need to prove that $a_n+1-a_n>0$ for all $ngeq 1$.

We would use induction on $n$.

For $n=1$, we have $a_2-a_1=sqrt2+sqrt2-sqrt2$

Now observe that, $sqrt2=(sqrt2+sqrt2)^2-(sqrt2)^2=(sqrt2+sqrt2-sqrt2)(sqrt2+sqrt2+sqrt2)$

And since $sqrt2>0$ and $sqrt2+sqrt2+sqrt2>0$, the above product shows that $sqrt2+sqrt2-sqrt2>0$ which is equivalent to $a_2-a_1>0$.

So the result is true for $n=1$.

Now assume that the result is true for $n=k$,i.e. $a_k+1>a_k$.

$a_k+2>a_k+1 iff sqrt2+a_k+1>a_k+1 iff 2+a_k+1>(a_k+1)^2 iff 2+a_k+1>2+a_k iff a_k+1>a_k$.

Thus, we have proved the result by induction.

share|cite|improve this answer

answered Mar 26 at 6:06

Ravi DwivediRavi Dwivedi

461

$endgroup$

add a comment | 

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1

$begingroup$

Hint: All quantities involved are positive, and $a_n+1=sqrt2+a_n>a_n$ if and only if $a_n^2<a_n+2$, i.e. $(a_n-2)(a_n+1)<0$. Can you put a bound on $a_n$ and hence use this inequality to prove the result?

share|cite|improve this answer

edited Mar 26 at 5:55

answered Mar 26 at 5:49

YiFanYiFan

5,3832728

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, an edit was made to the post that was actually incorrect. It is increasing. √2 < √(2+√2) and so on
  $endgroup$
  – jacksonf
  Mar 26 at 5:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have already proved that an is always less than 2. I have to say I'm still confused, how does this bound help prove the result?
  $endgroup$
  – jacksonf
  Mar 26 at 5:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $(a_n-2)(a_n+1)<0$ if and only if $a<2$ and $a_n>-1$. You have the first part, the second part is trivial. So the inequality is true!
  $endgroup$
  – YiFan
  Mar 26 at 6:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: All quantities involved are positive, and $a_n+1=sqrt2+a_n>a_n$ if and only if $a_n^2<a_n+2$, i.e. $(a_n-2)(a_n+1)<0$. Can you put a bound on $a_n$ and hence use this inequality to prove the result?

share|cite|improve this answer

edited Mar 26 at 5:55

answered Mar 26 at 5:49

YiFanYiFan

5,3832728

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, an edit was made to the post that was actually incorrect. It is increasing. √2 < √(2+√2) and so on
  $endgroup$
  – jacksonf
  Mar 26 at 5:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have already proved that an is always less than 2. I have to say I'm still confused, how does this bound help prove the result?
  $endgroup$
  – jacksonf
  Mar 26 at 5:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $(a_n-2)(a_n+1)<0$ if and only if $a<2$ and $a_n>-1$. You have the first part, the second part is trivial. So the inequality is true!
  $endgroup$
  – YiFan
  Mar 26 at 6:00
  

  
  
  
  

add a comment | 

$begingroup$

Hint: All quantities involved are positive, and $a_n+1=sqrt2+a_n>a_n$ if and only if $a_n^2<a_n+2$, i.e. $(a_n-2)(a_n+1)<0$. Can you put a bound on $a_n$ and hence use this inequality to prove the result?

share|cite|improve this answer

edited Mar 26 at 5:55

answered Mar 26 at 5:49

YiFanYiFan

5,3832728

$endgroup$

share|cite|improve this answer

edited Mar 26 at 5:55

answered Mar 26 at 5:49

YiFanYiFan

5,3832728

answered Mar 26 at 5:49

YiFanYiFan

5,3832728

answered Mar 26 at 5:49

YiFanYiFan

5,3832728

1

$begingroup$

First you check as an exercise that $a_n>0$ for all $ngeq 1$

We need to prove that $a_n+1-a_n>0$ for all $ngeq 1$.

We would use induction on $n$.

For $n=1$, we have $a_2-a_1=sqrt2+sqrt2-sqrt2$

Now observe that, $sqrt2=(sqrt2+sqrt2)^2-(sqrt2)^2=(sqrt2+sqrt2-sqrt2)(sqrt2+sqrt2+sqrt2)$

And since $sqrt2>0$ and $sqrt2+sqrt2+sqrt2>0$, the above product shows that $sqrt2+sqrt2-sqrt2>0$ which is equivalent to $a_2-a_1>0$.

So the result is true for $n=1$.

Now assume that the result is true for $n=k$,i.e. $a_k+1>a_k$.

$a_k+2>a_k+1 iff sqrt2+a_k+1>a_k+1 iff 2+a_k+1>(a_k+1)^2 iff 2+a_k+1>2+a_k iff a_k+1>a_k$.

Thus, we have proved the result by induction.

share|cite|improve this answer

answered Mar 26 at 6:06

Ravi DwivediRavi Dwivedi

461

$endgroup$

add a comment | 

1

$begingroup$

First you check as an exercise that $a_n>0$ for all $ngeq 1$

We need to prove that $a_n+1-a_n>0$ for all $ngeq 1$.

We would use induction on $n$.

For $n=1$, we have $a_2-a_1=sqrt2+sqrt2-sqrt2$

Now observe that, $sqrt2=(sqrt2+sqrt2)^2-(sqrt2)^2=(sqrt2+sqrt2-sqrt2)(sqrt2+sqrt2+sqrt2)$

And since $sqrt2>0$ and $sqrt2+sqrt2+sqrt2>0$, the above product shows that $sqrt2+sqrt2-sqrt2>0$ which is equivalent to $a_2-a_1>0$.

So the result is true for $n=1$.

Now assume that the result is true for $n=k$,i.e. $a_k+1>a_k$.

$a_k+2>a_k+1 iff sqrt2+a_k+1>a_k+1 iff 2+a_k+1>(a_k+1)^2 iff 2+a_k+1>2+a_k iff a_k+1>a_k$.

Thus, we have proved the result by induction.

share|cite|improve this answer

answered Mar 26 at 6:06

Ravi DwivediRavi Dwivedi

461

$endgroup$

add a comment | 

$begingroup$

First you check as an exercise that $a_n>0$ for all $ngeq 1$

We need to prove that $a_n+1-a_n>0$ for all $ngeq 1$.

We would use induction on $n$.

For $n=1$, we have $a_2-a_1=sqrt2+sqrt2-sqrt2$

Now observe that, $sqrt2=(sqrt2+sqrt2)^2-(sqrt2)^2=(sqrt2+sqrt2-sqrt2)(sqrt2+sqrt2+sqrt2)$

And since $sqrt2>0$ and $sqrt2+sqrt2+sqrt2>0$, the above product shows that $sqrt2+sqrt2-sqrt2>0$ which is equivalent to $a_2-a_1>0$.

So the result is true for $n=1$.

Now assume that the result is true for $n=k$,i.e. $a_k+1>a_k$.

$a_k+2>a_k+1 iff sqrt2+a_k+1>a_k+1 iff 2+a_k+1>(a_k+1)^2 iff 2+a_k+1>2+a_k iff a_k+1>a_k$.

Thus, we have proved the result by induction.

share|cite|improve this answer

answered Mar 26 at 6:06

Ravi DwivediRavi Dwivedi

461

$endgroup$

share|cite|improve this answer

answered Mar 26 at 6:06

Ravi DwivediRavi Dwivedi

461

answered Mar 26 at 6:06

Ravi DwivediRavi Dwivedi

461

answered Mar 26 at 6:06

Ravi DwivediRavi Dwivedi

461