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














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



















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












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











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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share|cite|improve this question




share|cite|improve this question









asked Jul 31 at 12:38









mushimaster

1399




1399











  • 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




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










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























  • 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










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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























  • 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














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























  • 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












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















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















share|cite|improve this answer



share|cite|improve this answer








edited Jul 31 at 13:49


























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
















  • 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












 

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