Sequence proof on absolute value

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So have to prove if $| a_n|to0$, then $a_kto0$ given that $a_k$ is a sequence.

I was wodnering how I might prove this, I considered that whats inside the absolute value must be positive and expanding it out, but got nowhere.

calculus proof-verification convergence

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asked Jul 31 at 12:38

mushimaster

1399

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just note that $| a_n | = | a_n - 0|$.
  – trancelocation
  Jul 31 at 12:42
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

So have to prove if $| a_n|to0$, then $a_kto0$ given that $a_k$ is a sequence.

I was wodnering how I might prove this, I considered that whats inside the absolute value must be positive and expanding it out, but got nowhere.

calculus proof-verification convergence

share|cite|improve this question

asked Jul 31 at 12:38

mushimaster

1399

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just note that $| a_n | = | a_n - 0|$.
  – trancelocation
  Jul 31 at 12:42
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

So have to prove if $| a_n|to0$, then $a_kto0$ given that $a_k$ is a sequence.

I was wodnering how I might prove this, I considered that whats inside the absolute value must be positive and expanding it out, but got nowhere.

calculus proof-verification convergence

share|cite|improve this question

asked Jul 31 at 12:38

mushimaster

1399

So have to prove if $| a_n|to0$, then $a_kto0$ given that $a_k$ is a sequence.

I was wodnering how I might prove this, I considered that whats inside the absolute value must be positive and expanding it out, but got nowhere.

calculus proof-verification convergence

share|cite|improve this question

asked Jul 31 at 12:38

mushimaster

1399

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 12:38

mushimaster

1399

asked Jul 31 at 12:38

mushimaster

1399

asked Jul 31 at 12:38

mushimaster

1399

1399

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just note that $| a_n | = | a_n - 0|$.
  – trancelocation
  Jul 31 at 12:42
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just note that $| a_n | = | a_n - 0|$.
  – trancelocation
  Jul 31 at 12:42
  

  
  
  
  

Just note that $| a_n | = | a_n - 0|$.
– trancelocation
Jul 31 at 12:42

Just note that $| a_n | = | a_n - 0|$.
– trancelocation
Jul 31 at 12:42

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

Using the $epsilon$-definition of the limit, you have that

$$forallepsilon >0:exists n_0inmathbbN:||a_n|-0|<epsilontext for all ngeq n_0$$

i.e., as $||a_n|-0|=||a_n||=|a_n|=|a_n-0|$, you have that

$$forallepsilon >0:exists n_0inmathbbN:|a_n-0|<epsilontext for all ngeq n_0$$

which is per definition $lim_ntoinftya_n=0$.

share|cite|improve this answer

edited Jul 31 at 13:49

answered Jul 31 at 12:46

zzuussee

1,172419

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's the difference between the $epsilon$-definition of the limit and the definition of the limit?
  – mathworker21
  Jul 31 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This almost went unnoticed by me, I think I just wanted to emphasize the chosen variant of stating the limit in this case, e.g. in difference to the definition in the context of topological spaces using neighborhoods(which is here equivalent, needless to say).
  – zzuussee
  Jul 31 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! is f.a mean for all?
  – mushimaster
  Jul 31 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I've edited my answer, as to not cause confusion in the future.
  – zzuussee
  Jul 31 at 13:48
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Using the $epsilon$-definition of the limit, you have that

$$forallepsilon >0:exists n_0inmathbbN:||a_n|-0|<epsilontext for all ngeq n_0$$

i.e., as $||a_n|-0|=||a_n||=|a_n|=|a_n-0|$, you have that

$$forallepsilon >0:exists n_0inmathbbN:|a_n-0|<epsilontext for all ngeq n_0$$

which is per definition $lim_ntoinftya_n=0$.

share|cite|improve this answer

edited Jul 31 at 13:49

answered Jul 31 at 12:46

zzuussee

1,172419

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's the difference between the $epsilon$-definition of the limit and the definition of the limit?
  – mathworker21
  Jul 31 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This almost went unnoticed by me, I think I just wanted to emphasize the chosen variant of stating the limit in this case, e.g. in difference to the definition in the context of topological spaces using neighborhoods(which is here equivalent, needless to say).
  – zzuussee
  Jul 31 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! is f.a mean for all?
  – mushimaster
  Jul 31 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I've edited my answer, as to not cause confusion in the future.
  – zzuussee
  Jul 31 at 13:48
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Using the $epsilon$-definition of the limit, you have that

$$forallepsilon >0:exists n_0inmathbbN:||a_n|-0|<epsilontext for all ngeq n_0$$

i.e., as $||a_n|-0|=||a_n||=|a_n|=|a_n-0|$, you have that

$$forallepsilon >0:exists n_0inmathbbN:|a_n-0|<epsilontext for all ngeq n_0$$

which is per definition $lim_ntoinftya_n=0$.

share|cite|improve this answer

edited Jul 31 at 13:49

answered Jul 31 at 12:46

zzuussee

1,172419

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's the difference between the $epsilon$-definition of the limit and the definition of the limit?
  – mathworker21
  Jul 31 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This almost went unnoticed by me, I think I just wanted to emphasize the chosen variant of stating the limit in this case, e.g. in difference to the definition in the context of topological spaces using neighborhoods(which is here equivalent, needless to say).
  – zzuussee
  Jul 31 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! is f.a mean for all?
  – mushimaster
  Jul 31 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I've edited my answer, as to not cause confusion in the future.
  – zzuussee
  Jul 31 at 13:48
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Using the $epsilon$-definition of the limit, you have that

$$forallepsilon >0:exists n_0inmathbbN:||a_n|-0|<epsilontext for all ngeq n_0$$

i.e., as $||a_n|-0|=||a_n||=|a_n|=|a_n-0|$, you have that

$$forallepsilon >0:exists n_0inmathbbN:|a_n-0|<epsilontext for all ngeq n_0$$

which is per definition $lim_ntoinftya_n=0$.

share|cite|improve this answer

edited Jul 31 at 13:49

answered Jul 31 at 12:46

zzuussee

1,172419

Using the $epsilon$-definition of the limit, you have that

$$forallepsilon >0:exists n_0inmathbbN:||a_n|-0|<epsilontext for all ngeq n_0$$

i.e., as $||a_n|-0|=||a_n||=|a_n|=|a_n-0|$, you have that

$$forallepsilon >0:exists n_0inmathbbN:|a_n-0|<epsilontext for all ngeq n_0$$

which is per definition $lim_ntoinftya_n=0$.

share|cite|improve this answer

edited Jul 31 at 13:49

answered Jul 31 at 12:46

zzuussee

1,172419

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 13:49

edited Jul 31 at 13:49

edited Jul 31 at 13:49

answered Jul 31 at 12:46

zzuussee

1,172419

answered Jul 31 at 12:46

zzuussee

1,172419

answered Jul 31 at 12:46

zzuussee

1,172419

1,172419

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's the difference between the $epsilon$-definition of the limit and the definition of the limit?
  – mathworker21
  Jul 31 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This almost went unnoticed by me, I think I just wanted to emphasize the chosen variant of stating the limit in this case, e.g. in difference to the definition in the context of topological spaces using neighborhoods(which is here equivalent, needless to say).
  – zzuussee
  Jul 31 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! is f.a mean for all?
  – mushimaster
  Jul 31 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I've edited my answer, as to not cause confusion in the future.
  – zzuussee
  Jul 31 at 13:48
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's the difference between the $epsilon$-definition of the limit and the definition of the limit?
  – mathworker21
  Jul 31 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This almost went unnoticed by me, I think I just wanted to emphasize the chosen variant of stating the limit in this case, e.g. in difference to the definition in the context of topological spaces using neighborhoods(which is here equivalent, needless to say).
  – zzuussee
  Jul 31 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks! is f.a mean for all?
  – mushimaster
  Jul 31 at 13:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes. I've edited my answer, as to not cause confusion in the future.
  – zzuussee
  Jul 31 at 13:48
  

  
  
  
  

What's the difference between the $epsilon$-definition of the limit and the definition of the limit?
– mathworker21
Jul 31 at 13:18

What's the difference between the $epsilon$-definition of the limit and the definition of the limit?
– mathworker21
Jul 31 at 13:18

This almost went unnoticed by me, I think I just wanted to emphasize the chosen variant of stating the limit in this case, e.g. in difference to the definition in the context of topological spaces using neighborhoods(which is here equivalent, needless to say).
– zzuussee
Jul 31 at 13:23

This almost went unnoticed by me, I think I just wanted to emphasize the chosen variant of stating the limit in this case, e.g. in difference to the definition in the context of topological spaces using neighborhoods(which is here equivalent, needless to say).
– zzuussee
Jul 31 at 13:23

thanks! is f.a mean for all?
– mushimaster
Jul 31 at 13:45

thanks! is f.a mean for all?
– mushimaster
Jul 31 at 13:45

Yes. I've edited my answer, as to not cause confusion in the future.
– zzuussee
Jul 31 at 13:48

Yes. I've edited my answer, as to not cause confusion in the future.
– zzuussee
Jul 31 at 13:48

add a comment | 

 

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