lim inf and lim sup of sequence obtained from combination of two sequences

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Suppose that $(a_n)_n$ and $(b_n)_n$ are two real sequences converging to $a$ and $b$ respectively (both in $mathbbR$). Let $a > b$ and define the sequence $(x_n)_n$ as follows:
$$x_2n = a_n quad x_2n-1 = b_n,$$
that is $(x_n)_n = (a_0, b_1, a_1, b_2, ldots)$.

I am asked to prove that lim inf of $x_n$ is $b$ and limsup equals $a$.

I first noticed that for some $n_0$ we have that $a_n > b_n$ for all $n geq n_0$. Indeed, the sequence $(a_n - b_n)_n$ converges to $a-b > 0$ and therefore there is an $n_0 in mathbbN$ such that
$$|a_n - b_n - (a-b)| < fraca-b2$$
which is equivalent with
$$fraca-b2 < a_n - b_n < 3fraca-b2.$$

In class, we defined lim inf as $lim_n to + infty y_n$ with $y_n = infx_k vert k geq n$. I think it suffices to look at $infb_k vert 2k-1 geq n$ (because $a_n > b_n$ for all $n$) to determine the $y_n$ (I still need to explain why I can discard of the $a_n$ in this infimum) and hence it limit, but I am stuck... The same with limsup. I know both exists, since the sequences $(a_n)_n$ and $(b_n)_n$ are both converging, hence bounded.

Can someone give a hint on this problem? Am I working in the right direction?

real-analysis sequences-and-series convergence limsup-and-liminf

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asked Jul 15 at 10:18

Student

1,9781626

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up vote
3
down vote

favorite

1

Suppose that $(a_n)_n$ and $(b_n)_n$ are two real sequences converging to $a$ and $b$ respectively (both in $mathbbR$). Let $a > b$ and define the sequence $(x_n)_n$ as follows:
$$x_2n = a_n quad x_2n-1 = b_n,$$
that is $(x_n)_n = (a_0, b_1, a_1, b_2, ldots)$.

I am asked to prove that lim inf of $x_n$ is $b$ and limsup equals $a$.

I first noticed that for some $n_0$ we have that $a_n > b_n$ for all $n geq n_0$. Indeed, the sequence $(a_n - b_n)_n$ converges to $a-b > 0$ and therefore there is an $n_0 in mathbbN$ such that
$$|a_n - b_n - (a-b)| < fraca-b2$$
which is equivalent with
$$fraca-b2 < a_n - b_n < 3fraca-b2.$$

In class, we defined lim inf as $lim_n to + infty y_n$ with $y_n = infx_k vert k geq n$. I think it suffices to look at $infb_k vert 2k-1 geq n$ (because $a_n > b_n$ for all $n$) to determine the $y_n$ (I still need to explain why I can discard of the $a_n$ in this infimum) and hence it limit, but I am stuck... The same with limsup. I know both exists, since the sequences $(a_n)_n$ and $(b_n)_n$ are both converging, hence bounded.

Can someone give a hint on this problem? Am I working in the right direction?

real-analysis sequences-and-series convergence limsup-and-liminf

share|cite|improve this question

asked Jul 15 at 10:18

Student

1,9781626

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

Suppose that $(a_n)_n$ and $(b_n)_n$ are two real sequences converging to $a$ and $b$ respectively (both in $mathbbR$). Let $a > b$ and define the sequence $(x_n)_n$ as follows:
$$x_2n = a_n quad x_2n-1 = b_n,$$
that is $(x_n)_n = (a_0, b_1, a_1, b_2, ldots)$.

I am asked to prove that lim inf of $x_n$ is $b$ and limsup equals $a$.

I first noticed that for some $n_0$ we have that $a_n > b_n$ for all $n geq n_0$. Indeed, the sequence $(a_n - b_n)_n$ converges to $a-b > 0$ and therefore there is an $n_0 in mathbbN$ such that
$$|a_n - b_n - (a-b)| < fraca-b2$$
which is equivalent with
$$fraca-b2 < a_n - b_n < 3fraca-b2.$$

In class, we defined lim inf as $lim_n to + infty y_n$ with $y_n = infx_k vert k geq n$. I think it suffices to look at $infb_k vert 2k-1 geq n$ (because $a_n > b_n$ for all $n$) to determine the $y_n$ (I still need to explain why I can discard of the $a_n$ in this infimum) and hence it limit, but I am stuck... The same with limsup. I know both exists, since the sequences $(a_n)_n$ and $(b_n)_n$ are both converging, hence bounded.

Can someone give a hint on this problem? Am I working in the right direction?

real-analysis sequences-and-series convergence limsup-and-liminf

share|cite|improve this question

asked Jul 15 at 10:18

Student

1,9781626

Suppose that $(a_n)_n$ and $(b_n)_n$ are two real sequences converging to $a$ and $b$ respectively (both in $mathbbR$). Let $a > b$ and define the sequence $(x_n)_n$ as follows:
$$x_2n = a_n quad x_2n-1 = b_n,$$
that is $(x_n)_n = (a_0, b_1, a_1, b_2, ldots)$.

I am asked to prove that lim inf of $x_n$ is $b$ and limsup equals $a$.

I first noticed that for some $n_0$ we have that $a_n > b_n$ for all $n geq n_0$. Indeed, the sequence $(a_n - b_n)_n$ converges to $a-b > 0$ and therefore there is an $n_0 in mathbbN$ such that
$$|a_n - b_n - (a-b)| < fraca-b2$$
which is equivalent with
$$fraca-b2 < a_n - b_n < 3fraca-b2.$$

In class, we defined lim inf as $lim_n to + infty y_n$ with $y_n = infx_k vert k geq n$. I think it suffices to look at $infb_k vert 2k-1 geq n$ (because $a_n > b_n$ for all $n$) to determine the $y_n$ (I still need to explain why I can discard of the $a_n$ in this infimum) and hence it limit, but I am stuck... The same with limsup. I know both exists, since the sequences $(a_n)_n$ and $(b_n)_n$ are both converging, hence bounded.

Can someone give a hint on this problem? Am I working in the right direction?

real-analysis sequences-and-series convergence limsup-and-liminf

share|cite|improve this question

asked Jul 15 at 10:18

Student

1,9781626

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 10:18

Student

1,9781626

asked Jul 15 at 10:18

Student

1,9781626

asked Jul 15 at 10:18

Student

1,9781626

1,9781626

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

Let $n_0 in mathbbN$ be such that $b_n le a_n, forall n ge n_0$.

For $n ge n_0$ we then have

$$infx_k : k ge n ge infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because $x_2k-1 le x_2k$.

On the other hand, in general we have

$$infx_k : k ge n le infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because we are taking the infimum over a smaller set.

Therefore $infx_k : k ge n = infb_k : 2k-1 ge n = infleftb_k : k ge fracn+12right$ for $n ge n_0$ so

$$liminf_ntoinfty x_n = lim_ntoinftyinfx_k : k ge n = lim_ntoinftyinfleftb_k : k ge fracn+12right = liminf_ntoinfty b_n = lim_ntoinfty b_n = b$$

The proof of $limsup_ntoinfty x_n = a$ is analogous. Also, you can consider the sequence $(-x_n)_n$ so the above proof shows

$$-a = liminf_ntoinfty -x_n = - limsup_ntoinfty x_n implies limsup_ntoinfty x_n = a$$

share|cite|improve this answer

edited Jul 15 at 10:37

answered Jul 15 at 10:33

mechanodroid

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The equality where you go from lim inf to the regular limit, how can we justify this?
  – Student
  Jul 15 at 10:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Student The sequence $(b_n)_n$ is convergent so $liminf_ntoinfty b_n = limsup_ntoinfty b_n = lim_ntoinfty b_n$.
  – mechanodroid
  Jul 15 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks!
  – Student
  Jul 15 at 10:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

If $c>a$, then $a_n,b_n<c$ if $n$ is large enough. Therefore, $x_n<c$ if $n$ is large enough and so $limsup_nx_nleqslant c$.

On the other hand, if $c<a$, then $a_n>c$ if $n$ is large enough and so $x_n>c$ infinitely often. Therefore, $limsup_nx_ngeqslant c$.

Since $c>aimplieslimsup_nx_nleqslant c$ and $c<aimplieslimsup_nx_ngeqslant c$, $limsup_nx_n=a$.

You can prove that $liminf_nx_n=b$ in a similar way.

share|cite|improve this answer

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer, I'll have to think about this when I'm home again tomorrow.
  – Student
  Jul 15 at 10:37
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Let $n_0 in mathbbN$ be such that $b_n le a_n, forall n ge n_0$.

For $n ge n_0$ we then have

$$infx_k : k ge n ge infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because $x_2k-1 le x_2k$.

On the other hand, in general we have

$$infx_k : k ge n le infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because we are taking the infimum over a smaller set.

Therefore $infx_k : k ge n = infb_k : 2k-1 ge n = infleftb_k : k ge fracn+12right$ for $n ge n_0$ so

$$liminf_ntoinfty x_n = lim_ntoinftyinfx_k : k ge n = lim_ntoinftyinfleftb_k : k ge fracn+12right = liminf_ntoinfty b_n = lim_ntoinfty b_n = b$$

The proof of $limsup_ntoinfty x_n = a$ is analogous. Also, you can consider the sequence $(-x_n)_n$ so the above proof shows

$$-a = liminf_ntoinfty -x_n = - limsup_ntoinfty x_n implies limsup_ntoinfty x_n = a$$

share|cite|improve this answer

edited Jul 15 at 10:37

answered Jul 15 at 10:33

mechanodroid

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The equality where you go from lim inf to the regular limit, how can we justify this?
  – Student
  Jul 15 at 10:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Student The sequence $(b_n)_n$ is convergent so $liminf_ntoinfty b_n = limsup_ntoinfty b_n = lim_ntoinfty b_n$.
  – mechanodroid
  Jul 15 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks!
  – Student
  Jul 15 at 10:39
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Let $n_0 in mathbbN$ be such that $b_n le a_n, forall n ge n_0$.

For $n ge n_0$ we then have

$$infx_k : k ge n ge infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because $x_2k-1 le x_2k$.

On the other hand, in general we have

$$infx_k : k ge n le infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because we are taking the infimum over a smaller set.

Therefore $infx_k : k ge n = infb_k : 2k-1 ge n = infleftb_k : k ge fracn+12right$ for $n ge n_0$ so

$$liminf_ntoinfty x_n = lim_ntoinftyinfx_k : k ge n = lim_ntoinftyinfleftb_k : k ge fracn+12right = liminf_ntoinfty b_n = lim_ntoinfty b_n = b$$

The proof of $limsup_ntoinfty x_n = a$ is analogous. Also, you can consider the sequence $(-x_n)_n$ so the above proof shows

$$-a = liminf_ntoinfty -x_n = - limsup_ntoinfty x_n implies limsup_ntoinfty x_n = a$$

share|cite|improve this answer

edited Jul 15 at 10:37

answered Jul 15 at 10:33

mechanodroid

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The equality where you go from lim inf to the regular limit, how can we justify this?
  – Student
  Jul 15 at 10:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Student The sequence $(b_n)_n$ is convergent so $liminf_ntoinfty b_n = limsup_ntoinfty b_n = lim_ntoinfty b_n$.
  – mechanodroid
  Jul 15 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks!
  – Student
  Jul 15 at 10:39
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Let $n_0 in mathbbN$ be such that $b_n le a_n, forall n ge n_0$.

For $n ge n_0$ we then have

$$infx_k : k ge n ge infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because $x_2k-1 le x_2k$.

On the other hand, in general we have

$$infx_k : k ge n le infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because we are taking the infimum over a smaller set.

Therefore $infx_k : k ge n = infb_k : 2k-1 ge n = infleftb_k : k ge fracn+12right$ for $n ge n_0$ so

$$liminf_ntoinfty x_n = lim_ntoinftyinfx_k : k ge n = lim_ntoinftyinfleftb_k : k ge fracn+12right = liminf_ntoinfty b_n = lim_ntoinfty b_n = b$$

The proof of $limsup_ntoinfty x_n = a$ is analogous. Also, you can consider the sequence $(-x_n)_n$ so the above proof shows

$$-a = liminf_ntoinfty -x_n = - limsup_ntoinfty x_n implies limsup_ntoinfty x_n = a$$

share|cite|improve this answer

edited Jul 15 at 10:37

answered Jul 15 at 10:33

mechanodroid

22.3k52041

Let $n_0 in mathbbN$ be such that $b_n le a_n, forall n ge n_0$.

For $n ge n_0$ we then have

$$infx_k : k ge n ge infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because $x_2k-1 le x_2k$.

On the other hand, in general we have

$$infx_k : k ge n le infx_2k-1 : 2k-1 ge n = infb_k : 2k-1 ge n$$

because we are taking the infimum over a smaller set.

Therefore $infx_k : k ge n = infb_k : 2k-1 ge n = infleftb_k : k ge fracn+12right$ for $n ge n_0$ so

$$liminf_ntoinfty x_n = lim_ntoinftyinfx_k : k ge n = lim_ntoinftyinfleftb_k : k ge fracn+12right = liminf_ntoinfty b_n = lim_ntoinfty b_n = b$$

The proof of $limsup_ntoinfty x_n = a$ is analogous. Also, you can consider the sequence $(-x_n)_n$ so the above proof shows

$$-a = liminf_ntoinfty -x_n = - limsup_ntoinfty x_n implies limsup_ntoinfty x_n = a$$

share|cite|improve this answer

edited Jul 15 at 10:37

answered Jul 15 at 10:33

mechanodroid

22.3k52041

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 10:37

edited Jul 15 at 10:37

edited Jul 15 at 10:37

answered Jul 15 at 10:33

mechanodroid

22.3k52041

answered Jul 15 at 10:33

mechanodroid

22.3k52041

answered Jul 15 at 10:33

mechanodroid

22.3k52041

22.3k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The equality where you go from lim inf to the regular limit, how can we justify this?
  – Student
  Jul 15 at 10:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Student The sequence $(b_n)_n$ is convergent so $liminf_ntoinfty b_n = limsup_ntoinfty b_n = lim_ntoinfty b_n$.
  – mechanodroid
  Jul 15 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks!
  – Student
  Jul 15 at 10:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The equality where you go from lim inf to the regular limit, how can we justify this?
  – Student
  Jul 15 at 10:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Student The sequence $(b_n)_n$ is convergent so $liminf_ntoinfty b_n = limsup_ntoinfty b_n = lim_ntoinfty b_n$.
  – mechanodroid
  Jul 15 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks!
  – Student
  Jul 15 at 10:39
  

  
  
  
  

The equality where you go from lim inf to the regular limit, how can we justify this?
– Student
Jul 15 at 10:36

The equality where you go from lim inf to the regular limit, how can we justify this?
– Student
Jul 15 at 10:36

@Student The sequence $(b_n)_n$ is convergent so $liminf_ntoinfty b_n = limsup_ntoinfty b_n = lim_ntoinfty b_n$.
– mechanodroid
Jul 15 at 10:38

@Student The sequence $(b_n)_n$ is convergent so $liminf_ntoinfty b_n = limsup_ntoinfty b_n = lim_ntoinfty b_n$.
– mechanodroid
Jul 15 at 10:38

Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks!
– Student
Jul 15 at 10:39

Oh yes, stupid of me. I'll read your answer more carefully when I'm back home tomorrow. Thanks!
– Student
Jul 15 at 10:39

add a comment | 

up vote
1
down vote

If $c>a$, then $a_n,b_n<c$ if $n$ is large enough. Therefore, $x_n<c$ if $n$ is large enough and so $limsup_nx_nleqslant c$.

On the other hand, if $c<a$, then $a_n>c$ if $n$ is large enough and so $x_n>c$ infinitely often. Therefore, $limsup_nx_ngeqslant c$.

Since $c>aimplieslimsup_nx_nleqslant c$ and $c<aimplieslimsup_nx_ngeqslant c$, $limsup_nx_n=a$.

You can prove that $liminf_nx_n=b$ in a similar way.

share|cite|improve this answer

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer, I'll have to think about this when I'm home again tomorrow.
  – Student
  Jul 15 at 10:37
  

  
  
  
  

add a comment | 

up vote
1
down vote

If $c>a$, then $a_n,b_n<c$ if $n$ is large enough. Therefore, $x_n<c$ if $n$ is large enough and so $limsup_nx_nleqslant c$.

On the other hand, if $c<a$, then $a_n>c$ if $n$ is large enough and so $x_n>c$ infinitely often. Therefore, $limsup_nx_ngeqslant c$.

Since $c>aimplieslimsup_nx_nleqslant c$ and $c<aimplieslimsup_nx_ngeqslant c$, $limsup_nx_n=a$.

You can prove that $liminf_nx_n=b$ in a similar way.

share|cite|improve this answer

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer, I'll have to think about this when I'm home again tomorrow.
  – Student
  Jul 15 at 10:37
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

If $c>a$, then $a_n,b_n<c$ if $n$ is large enough. Therefore, $x_n<c$ if $n$ is large enough and so $limsup_nx_nleqslant c$.

On the other hand, if $c<a$, then $a_n>c$ if $n$ is large enough and so $x_n>c$ infinitely often. Therefore, $limsup_nx_ngeqslant c$.

Since $c>aimplieslimsup_nx_nleqslant c$ and $c<aimplieslimsup_nx_ngeqslant c$, $limsup_nx_n=a$.

You can prove that $liminf_nx_n=b$ in a similar way.

share|cite|improve this answer

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

If $c>a$, then $a_n,b_n<c$ if $n$ is large enough. Therefore, $x_n<c$ if $n$ is large enough and so $limsup_nx_nleqslant c$.

On the other hand, if $c<a$, then $a_n>c$ if $n$ is large enough and so $x_n>c$ infinitely often. Therefore, $limsup_nx_ngeqslant c$.

Since $c>aimplieslimsup_nx_nleqslant c$ and $c<aimplieslimsup_nx_ngeqslant c$, $limsup_nx_n=a$.

You can prove that $liminf_nx_n=b$ in a similar way.

share|cite|improve this answer

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

answered Jul 15 at 10:25

José Carlos Santos

114k1698177

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer, I'll have to think about this when I'm home again tomorrow.
  – Student
  Jul 15 at 10:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer, I'll have to think about this when I'm home again tomorrow.
  – Student
  Jul 15 at 10:37
  

  
  
  
  

Thanks for your answer, I'll have to think about this when I'm home again tomorrow.
– Student
Jul 15 at 10:37

Thanks for your answer, I'll have to think about this when I'm home again tomorrow.
– Student
Jul 15 at 10:37

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