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A sequence that changes sign finally at infinity?

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1

$begingroup$

I want to make a sequence $a_n$ that changes its sign at infinity.

What I'm try to make is this: $a_n$ of negative number or zero $to$ it's limit is some positive $epsilon$ right next to zero.

$qquad a_n le 0$ (for all n $in Z^+$)

$qquad limlimits_n to infty a_n = epsilon $, where $epsilongt 0$.

Is this sequence possible?

Could you find an explicit function?
$,,$ Thanks in advance.

real-analysis sequences-and-series limits analysis

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

asked 9 hours ago

yoshiyoshi

822 bronze badges

New contributor

yoshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is impossible. Try considering the epsilon-delta definition of a limit.
  $endgroup$
  – Peter Foreman
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no, it's not possible
  $endgroup$
  – J. W. Tanner
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not possible, sorry. $0$ is an upper bound for your entire sequence so $limsup_nto inftya_n≤0$.
  $endgroup$
  – lulu
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I want to make a sequence $a_n$ that changes its sign at infinity.

What I'm try to make is this: $a_n$ of negative number or zero $to$ it's limit is some positive $epsilon$ right next to zero.

$qquad a_n le 0$ (for all n $in Z^+$)

$qquad limlimits_n to infty a_n = epsilon $, where $epsilongt 0$.

Is this sequence possible?

Could you find an explicit function?
$,,$ Thanks in advance.

real-analysis sequences-and-series limits analysis

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

asked 9 hours ago

yoshiyoshi

822 bronze badges

New contributor

yoshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is impossible. Try considering the epsilon-delta definition of a limit.
  $endgroup$
  – Peter Foreman
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no, it's not possible
  $endgroup$
  – J. W. Tanner
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not possible, sorry. $0$ is an upper bound for your entire sequence so $limsup_nto inftya_n≤0$.
  $endgroup$
  – lulu
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I want to make a sequence $a_n$ that changes its sign at infinity.

What I'm try to make is this: $a_n$ of negative number or zero $to$ it's limit is some positive $epsilon$ right next to zero.

$qquad a_n le 0$ (for all n $in Z^+$)

$qquad limlimits_n to infty a_n = epsilon $, where $epsilongt 0$.

Is this sequence possible?

Could you find an explicit function?
$,,$ Thanks in advance.

real-analysis sequences-and-series limits analysis

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

asked 9 hours ago

yoshiyoshi

822 bronze badges

New contributor

yoshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

asked 9 hours ago

yoshiyoshi

822 bronze badges

New contributor

yoshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 8 hours ago

José Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

asked 9 hours ago

yoshiyoshi

822 bronze badges

New contributor

yoshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 8 hours ago

José Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

edited 8 hours ago

José Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

asked 9 hours ago

yoshiyoshi

822 bronze badges

New contributor

yoshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

yoshiyoshi

822 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

No, it is not possible. Suppose otherwise. Then, there is some $Ninmathbb N$ such that$$ngeqslant Nimplieslvert a_n-varepsilonrvert<varepsiloniff0<a_n<2varepsilon.$$But this is impossible, since you are assuming that you always have $a_nleqslant0$.

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Such a sequence doesn't exist. Note that if $a_n le 0$ for all $ninBbb N_0$, then $0$ is an upper bound of the sequence, so in particular

$$lim_ntoinfty a_n le limsup_ntoinfty a_nlesup_ninBbb N_0 a_n le 0.$$

share|cite|improve this answer

answered 8 hours ago

Maximilian JanischMaximilian Janisch

1,90944 silver badges2020 bronze badges

$endgroup$

add a comment | 

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6

$begingroup$

No, it is not possible. Suppose otherwise. Then, there is some $Ninmathbb N$ such that$$ngeqslant Nimplieslvert a_n-varepsilonrvert<varepsiloniff0<a_n<2varepsilon.$$But this is impossible, since you are assuming that you always have $a_nleqslant0$.

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

No, it is not possible. Suppose otherwise. Then, there is some $Ninmathbb N$ such that$$ngeqslant Nimplieslvert a_n-varepsilonrvert<varepsiloniff0<a_n<2varepsilon.$$But this is impossible, since you are assuming that you always have $a_nleqslant0$.

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

$endgroup$

add a comment | 

$begingroup$

No, it is not possible. Suppose otherwise. Then, there is some $Ninmathbb N$ such that$$ngeqslant Nimplieslvert a_n-varepsilonrvert<varepsiloniff0<a_n<2varepsilon.$$But this is impossible, since you are assuming that you always have $a_nleqslant0$.

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

198k2424 gold badges156156 silver badges273273 bronze badges

2

$begingroup$

Such a sequence doesn't exist. Note that if $a_n le 0$ for all $ninBbb N_0$, then $0$ is an upper bound of the sequence, so in particular

$$lim_ntoinfty a_n le limsup_ntoinfty a_nlesup_ninBbb N_0 a_n le 0.$$

share|cite|improve this answer

answered 8 hours ago

Maximilian JanischMaximilian Janisch

1,90944 silver badges2020 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Such a sequence doesn't exist. Note that if $a_n le 0$ for all $ninBbb N_0$, then $0$ is an upper bound of the sequence, so in particular

$$lim_ntoinfty a_n le limsup_ntoinfty a_nlesup_ninBbb N_0 a_n le 0.$$

share|cite|improve this answer

answered 8 hours ago

Maximilian JanischMaximilian Janisch

1,90944 silver badges2020 bronze badges

$endgroup$

add a comment | 

$begingroup$

Such a sequence doesn't exist. Note that if $a_n le 0$ for all $ninBbb N_0$, then $0$ is an upper bound of the sequence, so in particular

$$lim_ntoinfty a_n le limsup_ntoinfty a_nlesup_ninBbb N_0 a_n le 0.$$

share|cite|improve this answer

answered 8 hours ago

Maximilian JanischMaximilian Janisch

1,90944 silver badges2020 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Maximilian JanischMaximilian Janisch

1,90944 silver badges2020 bronze badges

answered 8 hours ago

Maximilian JanischMaximilian Janisch

1,90944 silver badges2020 bronze badges

answered 8 hours ago

Maximilian JanischMaximilian Janisch

1,90944 silver badges2020 bronze badges