Sequence Converging [on hold]
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Explain exactly what it means for $a_n$ $ninmathbb N$ to converge to $L ∈ R.$
Can someone please break down this question for me?
calculus sequences-and-series limits
put on hold as off-topic by Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128
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up vote
-3
down vote
favorite
Explain exactly what it means for $a_n$ $ninmathbb N$ to converge to $L ∈ R.$
Can someone please break down this question for me?
calculus sequences-and-series limits
put on hold as off-topic by Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128
5
Can you edit your question to include the definition you are using, and what you don't understand about the definition?
– quasi
2 days ago
I'm voting to close this question until quasi's comment receives an appropriate answer.
– Arnaud Mortier
2 days ago
Please study our guide to new askers. You shoulld then understand why your question was not well received, and can take action.
– Jyrki Lahtonen
2 days ago
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up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
Explain exactly what it means for $a_n$ $ninmathbb N$ to converge to $L ∈ R.$
Can someone please break down this question for me?
calculus sequences-and-series limits
Explain exactly what it means for $a_n$ $ninmathbb N$ to converge to $L ∈ R.$
Can someone please break down this question for me?
calculus sequences-and-series limits
edited 2 days ago
Michael Hardy
204k23185460
204k23185460
asked 2 days ago


Okie
395
395
put on hold as off-topic by Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128
put on hold as off-topic by Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, José Carlos Santos, amWhy, Arnaud Mortier, user 108128
5
Can you edit your question to include the definition you are using, and what you don't understand about the definition?
– quasi
2 days ago
I'm voting to close this question until quasi's comment receives an appropriate answer.
– Arnaud Mortier
2 days ago
Please study our guide to new askers. You shoulld then understand why your question was not well received, and can take action.
– Jyrki Lahtonen
2 days ago
add a comment |Â
5
Can you edit your question to include the definition you are using, and what you don't understand about the definition?
– quasi
2 days ago
I'm voting to close this question until quasi's comment receives an appropriate answer.
– Arnaud Mortier
2 days ago
Please study our guide to new askers. You shoulld then understand why your question was not well received, and can take action.
– Jyrki Lahtonen
2 days ago
5
5
Can you edit your question to include the definition you are using, and what you don't understand about the definition?
– quasi
2 days ago
Can you edit your question to include the definition you are using, and what you don't understand about the definition?
– quasi
2 days ago
I'm voting to close this question until quasi's comment receives an appropriate answer.
– Arnaud Mortier
2 days ago
I'm voting to close this question until quasi's comment receives an appropriate answer.
– Arnaud Mortier
2 days ago
Please study our guide to new askers. You shoulld then understand why your question was not well received, and can take action.
– Jyrki Lahtonen
2 days ago
Please study our guide to new askers. You shoulld then understand why your question was not well received, and can take action.
– Jyrki Lahtonen
2 days ago
add a comment |Â
1 Answer
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The definition says the following:
$forallepsilon>0$ $exists n_0in mathbbN$ $forall ngeq n_0$ $[|a_n-L|<epsilon]$
Take any positive number $epsilon$, doesn't matter how small it is. There must be an index $n_0 in mathbbN$ that starting from it the distance of all of the elements in the sequence from $L$ is smaller than $epsilon$. Of course the index $n_0$ depends on $epsilon$.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The definition says the following:
$forallepsilon>0$ $exists n_0in mathbbN$ $forall ngeq n_0$ $[|a_n-L|<epsilon]$
Take any positive number $epsilon$, doesn't matter how small it is. There must be an index $n_0 in mathbbN$ that starting from it the distance of all of the elements in the sequence from $L$ is smaller than $epsilon$. Of course the index $n_0$ depends on $epsilon$.
add a comment |Â
up vote
0
down vote
The definition says the following:
$forallepsilon>0$ $exists n_0in mathbbN$ $forall ngeq n_0$ $[|a_n-L|<epsilon]$
Take any positive number $epsilon$, doesn't matter how small it is. There must be an index $n_0 in mathbbN$ that starting from it the distance of all of the elements in the sequence from $L$ is smaller than $epsilon$. Of course the index $n_0$ depends on $epsilon$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The definition says the following:
$forallepsilon>0$ $exists n_0in mathbbN$ $forall ngeq n_0$ $[|a_n-L|<epsilon]$
Take any positive number $epsilon$, doesn't matter how small it is. There must be an index $n_0 in mathbbN$ that starting from it the distance of all of the elements in the sequence from $L$ is smaller than $epsilon$. Of course the index $n_0$ depends on $epsilon$.
The definition says the following:
$forallepsilon>0$ $exists n_0in mathbbN$ $forall ngeq n_0$ $[|a_n-L|<epsilon]$
Take any positive number $epsilon$, doesn't matter how small it is. There must be an index $n_0 in mathbbN$ that starting from it the distance of all of the elements in the sequence from $L$ is smaller than $epsilon$. Of course the index $n_0$ depends on $epsilon$.
answered 2 days ago
Mark
5849
5849
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5
Can you edit your question to include the definition you are using, and what you don't understand about the definition?
– quasi
2 days ago
I'm voting to close this question until quasi's comment receives an appropriate answer.
– Arnaud Mortier
2 days ago
Please study our guide to new askers. You shoulld then understand why your question was not well received, and can take action.
– Jyrki Lahtonen
2 days ago