Mixing
Should the formal definition of the limit of a sequence be improved?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Consider the "formal definition" here https://en.wikipedia.org/wiki/Limit_of_a_sequence. I checked some references and this is often precisely the definition in all words and terms used in this article. I claim that this definition is not precise or incomplete. Probably I am wrong and that is why I would like to ask about it. Consider the "illustration" part. This figure makes the concept very clear. It shows that as $varepsilon$ decreases $N$ should increase. So this means that (i) $N$ must be a function of $varepsilon$, and (ii) $N$ must be a decreasing function of $varepsilon$. Why are these not made part of the formal definition by at least writing $N(varepsilon)$ instead of $N$ and perhaps adding a sentence that $N$ must be a decreasing function of $varepsilon$? If these are not made explicit, why is it clear from the formal definition that $x_n$ actually converges to $x$ and $n$ increases?
sequences-and-series limits definition
asked 1 hour ago
Snoopy
233
 |Â
show 4 more comments
up vote
2
down vote
favorite
Consider the "formal definition" here https://en.wikipedia.org/wiki/Limit_of_a_sequence. I checked some references and this is often precisely the definition in all words and terms used in this article. I claim that this definition is not precise or incomplete. Probably I am wrong and that is why I would like to ask about it. Consider the "illustration" part. This figure makes the concept very clear. It shows that as $varepsilon$ decreases $N$ should increase. So this means that (i) $N$ must be a function of $varepsilon$, and (ii) $N$ must be a decreasing function of $varepsilon$. Why are these not made part of the formal definition by at least writing $N(varepsilon)$ instead of $N$ and perhaps adding a sentence that $N$ must be a decreasing function of $varepsilon$? If these are not made explicit, why is it clear from the formal definition that $x_n$ actually converges to $x$ and $n$ increases?
sequences-and-series limits definition
asked 1 hour ago
Snoopy
233
1
$N$ is never unique.
– Randall
1 hour ago
Also consider constant sequences.
– Randall
1 hour ago
1
Perhaps you are thinking of $N$ as the minimal integer such that $|x_n - x|< epsilon$ for $n ge N$. But that is not a practical definition.
– Martin R
1 hour ago
$N$ would also depend on the sequence, so $N(epsilon, a_1,a_2,ldots)$? Seriously, we do not use functional notation in this place, the correct nesting of quantors is enough (and works with the common rules of inference involving them)
– Hagen von Eitzen
1 hour ago
@MartinR, this is indeed what I am at. Why is it not a practical definition? This is what the figure of the article says right? As $N$ increases, $varepsilon$ must decrease, right? Otherwise convergence does not take place.
– Snoopy
1 hour ago
 |Â
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the "formal definition" here https://en.wikipedia.org/wiki/Limit_of_a_sequence. I checked some references and this is often precisely the definition in all words and terms used in this article. I claim that this definition is not precise or incomplete. Probably I am wrong and that is why I would like to ask about it. Consider the "illustration" part. This figure makes the concept very clear. It shows that as $varepsilon$ decreases $N$ should increase. So this means that (i) $N$ must be a function of $varepsilon$, and (ii) $N$ must be a decreasing function of $varepsilon$. Why are these not made part of the formal definition by at least writing $N(varepsilon)$ instead of $N$ and perhaps adding a sentence that $N$ must be a decreasing function of $varepsilon$? If these are not made explicit, why is it clear from the formal definition that $x_n$ actually converges to $x$ and $n$ increases?
sequences-and-series limits definition
asked 1 hour ago
Snoopy
233
Consider the "formal definition" here https://en.wikipedia.org/wiki/Limit_of_a_sequence. I checked some references and this is often precisely the definition in all words and terms used in this article. I claim that this definition is not precise or incomplete. Probably I am wrong and that is why I would like to ask about it. Consider the "illustration" part. This figure makes the concept very clear. It shows that as $varepsilon$ decreases $N$ should increase. So this means that (i) $N$ must be a function of $varepsilon$, and (ii) $N$ must be a decreasing function of $varepsilon$. Why are these not made part of the formal definition by at least writing $N(varepsilon)$ instead of $N$ and perhaps adding a sentence that $N$ must be a decreasing function of $varepsilon$? If these are not made explicit, why is it clear from the formal definition that $x_n$ actually converges to $x$ and $n$ increases?
sequences-and-series limits definition
sequences-and-series limits definition
asked 1 hour ago
Snoopy
233
asked 1 hour ago
Snoopy
233
asked 1 hour ago
Snoopy
233
asked 1 hour ago
Snoopy
233
asked 1 hour ago
Snoopy
233
233
1
$N$ is never unique.
– Randall
1 hour ago
Also consider constant sequences.
– Randall
1 hour ago
1
Perhaps you are thinking of $N$ as the minimal integer such that $|x_n - x|< epsilon$ for $n ge N$. But that is not a practical definition.
– Martin R
1 hour ago
$N$ would also depend on the sequence, so $N(epsilon, a_1,a_2,ldots)$? Seriously, we do not use functional notation in this place, the correct nesting of quantors is enough (and works with the common rules of inference involving them)
– Hagen von Eitzen
1 hour ago
@MartinR, this is indeed what I am at. Why is it not a practical definition? This is what the figure of the article says right? As $N$ increases, $varepsilon$ must decrease, right? Otherwise convergence does not take place.
– Snoopy
1 hour ago
 |Â
show 4 more comments
1
$N$ is never unique.
– Randall
1 hour ago
Also consider constant sequences.
– Randall
1 hour ago
1
Perhaps you are thinking of $N$ as the minimal integer such that $|x_n - x|< epsilon$ for $n ge N$. But that is not a practical definition.
– Martin R
1 hour ago
$N$ would also depend on the sequence, so $N(epsilon, a_1,a_2,ldots)$? Seriously, we do not use functional notation in this place, the correct nesting of quantors is enough (and works with the common rules of inference involving them)
– Hagen von Eitzen
1 hour ago
@MartinR, this is indeed what I am at. Why is it not a practical definition? This is what the figure of the article says right? As $N$ increases, $varepsilon$ must decrease, right? Otherwise convergence does not take place.
– Snoopy
1 hour ago
1
1
$N$ is never unique.
– Randall
1 hour ago
$N$ is never unique.
– Randall
1 hour ago
Also consider constant sequences.
– Randall
1 hour ago
Also consider constant sequences.
– Randall
1 hour ago
1
1
Perhaps you are thinking of $N$ as the minimal integer such that $|x_n - x|< epsilon$ for $n ge N$. But that is not a practical definition.
– Martin R
1 hour ago
Perhaps you are thinking of $N$ as the minimal integer such that $|x_n - x|< epsilon$ for $n ge N$. But that is not a practical definition.
– Martin R
1 hour ago
$N$ would also depend on the sequence, so $N(epsilon, a_1,a_2,ldots)$? Seriously, we do not use functional notation in this place, the correct nesting of quantors is enough (and works with the common rules of inference involving them)
– Hagen von Eitzen
1 hour ago
$N$ would also depend on the sequence, so $N(epsilon, a_1,a_2,ldots)$? Seriously, we do not use functional notation in this place, the correct nesting of quantors is enough (and works with the common rules of inference involving them)
– Hagen von Eitzen
1 hour ago
@MartinR, this is indeed what I am at. Why is it not a practical definition? This is what the figure of the article says right? As $N$ increases, $varepsilon$ must decrease, right? Otherwise convergence does not take place.
– Snoopy
1 hour ago
@MartinR, this is indeed what I am at. Why is it not a practical definition? This is what the figure of the article says right? As $N$ increases, $varepsilon$ must decrease, right? Otherwise convergence does not take place.
– Snoopy
1 hour ago
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
$N$ need not be a function of $epsilon$. For a given $epsilon$, many $N$ are possible (in particular all values larger than $N$ also work).
Enforcing $N$ to be some function a $epsilon$ wouldn't be a good idea, because it would impose a constructive solution, i.e. require you to exhibit the particular value of $N(epsilon)$. Though in reality, it suffices to prove that a suitable value exists, whatever it is.
answered 1 hour ago
Yves Daoust
118k667213
Also, having "$N$ must be a decreasing function of $varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds.
– Dave L. Renfro
1 hour ago
@Yves (1) Why is it true that if I decrease $varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $varepsilon$ and $N$ should have a link. Right?
– Snoopy
56 mins ago
To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version.
– Yves Daoust
28 mins ago
add a comment |Â
up vote
2
down vote
You are right that $N$ depends on $epsilon$ - and missing this fact is the cause of a lot of confusion among people learning analysis for the first time. Sometimes it is made explicit: "for all $epsilon > 0$ there exists an $N$ (dependent on $epsilon$) such that...". But even when phrased as "for all $epsilon > 0$ there exists an $N$ such that...", $N$ still depends on $epsilon$: that's how quantifiers ("for all", "there exists") work, even if it's not made explicit.
One thing that should be pointed out is that $N$ is not unique. Indeed if, for a given $epsilon$, $N = M$ works[1], then $N = M'$ works for any $M' geq M$. We could define things such that $N$ was the minimum that worked, but this would make both the definition and proofs involving the definition more convoluted, with very little gain in return.
With this in mind, it's somewhat unhelpful to think of $N$ as a function of $epsilon$, because "function" suggests there's a unique value of $N$ for each $epsilon$, which isn't true.
On the topic of $N$ increasing as $epsilon$ decreases: this is not true in the most pedantic sense: if $x_i = 0$ for all $i$, then we could take $N = 10$ when $epsilon > 1$ and $N = 1$ when $epsilon leq 1$, the definition is satisfied but $N$ is not increasing as $epsilon$ decreases.
However, in a less pedantic sense, it is true that if $N$ works for $epsilon$ and $epsilon' > epsilon$ then $N$ works for $epsilon'$ too. But this is a consequence of the definition, and so doesn't need to be included in the definition.
To answer your third question, "why is it clear from the formal definition that $x_n$ actually converges to $x$ as $n$ increases?", there's a temptation to ask "what do you mean by 'actually converges'?" Perhaps a helpful observation is that the definition is equivalent to saying: for any $epsilon > 0$, all but finitely many of the terms of the sequence are within $epsilon$ of $x$. We don't care how many that "finitely many" is, provided there are only finitely many of them.
[1]: I hope it's clear what I mean by "$N$ works for $epsilon$" here and later, I just don't want to have to keep writing out the second half of the definition
answered 50 mins ago
Christopher
5,44911526
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$N$ need not be a function of $epsilon$. For a given $epsilon$, many $N$ are possible (in particular all values larger than $N$ also work).
Enforcing $N$ to be some function a $epsilon$ wouldn't be a good idea, because it would impose a constructive solution, i.e. require you to exhibit the particular value of $N(epsilon)$. Though in reality, it suffices to prove that a suitable value exists, whatever it is.
answered 1 hour ago
Yves Daoust
118k667213
Also, having "$N$ must be a decreasing function of $varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds.
– Dave L. Renfro
1 hour ago
@Yves (1) Why is it true that if I decrease $varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $varepsilon$ and $N$ should have a link. Right?
– Snoopy
56 mins ago
To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version.
– Yves Daoust
28 mins ago
add a comment |Â
up vote
3
down vote
$N$ need not be a function of $epsilon$. For a given $epsilon$, many $N$ are possible (in particular all values larger than $N$ also work).
Enforcing $N$ to be some function a $epsilon$ wouldn't be a good idea, because it would impose a constructive solution, i.e. require you to exhibit the particular value of $N(epsilon)$. Though in reality, it suffices to prove that a suitable value exists, whatever it is.
answered 1 hour ago
Yves Daoust
118k667213
Also, having "$N$ must be a decreasing function of $varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds.
– Dave L. Renfro
1 hour ago
@Yves (1) Why is it true that if I decrease $varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $varepsilon$ and $N$ should have a link. Right?
– Snoopy
56 mins ago
To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version.
– Yves Daoust
28 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$N$ need not be a function of $epsilon$. For a given $epsilon$, many $N$ are possible (in particular all values larger than $N$ also work).
Enforcing $N$ to be some function a $epsilon$ wouldn't be a good idea, because it would impose a constructive solution, i.e. require you to exhibit the particular value of $N(epsilon)$. Though in reality, it suffices to prove that a suitable value exists, whatever it is.
answered 1 hour ago
Yves Daoust
118k667213
$N$ need not be a function of $epsilon$. For a given $epsilon$, many $N$ are possible (in particular all values larger than $N$ also work).
Enforcing $N$ to be some function a $epsilon$ wouldn't be a good idea, because it would impose a constructive solution, i.e. require you to exhibit the particular value of $N(epsilon)$. Though in reality, it suffices to prove that a suitable value exists, whatever it is.
answered 1 hour ago
Yves Daoust
118k667213
edited 33 mins ago
answered 1 hour ago
Yves Daoust
118k667213
answered 1 hour ago
Yves Daoust
118k667213
answered 1 hour ago
Yves Daoust
118k667213
118k667213
Also, having "$N$ must be a decreasing function of $varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds.
– Dave L. Renfro
1 hour ago
@Yves (1) Why is it true that if I decrease $varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $varepsilon$ and $N$ should have a link. Right?
– Snoopy
56 mins ago
To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version.
– Yves Daoust
28 mins ago
add a comment |Â
Also, having "$N$ must be a decreasing function of $varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds.
– Dave L. Renfro
1 hour ago
@Yves (1) Why is it true that if I decrease $varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $varepsilon$ and $N$ should have a link. Right?
– Snoopy
56 mins ago
To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version.
– Yves Daoust
28 mins ago
Also, having "$N$ must be a decreasing function of $varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds.
– Dave L. Renfro
1 hour ago
Also, having "$N$ must be a decreasing function of $varepsilon$" as part of the definition means this will have to be verified in proofs, or reference will have to be made to a(n unnecessary) theorem that says without this assumption the result still holds.
– Dave L. Renfro
1 hour ago
@Yves (1) Why is it true that if I decrease $varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $varepsilon$ and $N$ should have a link. Right?
– Snoopy
56 mins ago
@Yves (1) Why is it true that if I decrease $varepsilon$ I necessarily get a subset of the possible $N$? (2) This claim also makes it clear that $varepsilon$ and $N$ should have a link. Right?
– Snoopy
56 mins ago
To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version.
– Yves Daoust
28 mins ago
To the upvoters: I thank you, but I have completely rewritten since. Feel free to unupvote if you don't like the new version.
– Yves Daoust
28 mins ago
add a comment |Â
up vote
2
down vote
You are right that $N$ depends on $epsilon$ - and missing this fact is the cause of a lot of confusion among people learning analysis for the first time. Sometimes it is made explicit: "for all $epsilon > 0$ there exists an $N$ (dependent on $epsilon$) such that...". But even when phrased as "for all $epsilon > 0$ there exists an $N$ such that...", $N$ still depends on $epsilon$: that's how quantifiers ("for all", "there exists") work, even if it's not made explicit.
One thing that should be pointed out is that $N$ is not unique. Indeed if, for a given $epsilon$, $N = M$ works[1], then $N = M'$ works for any $M' geq M$. We could define things such that $N$ was the minimum that worked, but this would make both the definition and proofs involving the definition more convoluted, with very little gain in return.
With this in mind, it's somewhat unhelpful to think of $N$ as a function of $epsilon$, because "function" suggests there's a unique value of $N$ for each $epsilon$, which isn't true.
On the topic of $N$ increasing as $epsilon$ decreases: this is not true in the most pedantic sense: if $x_i = 0$ for all $i$, then we could take $N = 10$ when $epsilon > 1$ and $N = 1$ when $epsilon leq 1$, the definition is satisfied but $N$ is not increasing as $epsilon$ decreases.
However, in a less pedantic sense, it is true that if $N$ works for $epsilon$ and $epsilon' > epsilon$ then $N$ works for $epsilon'$ too. But this is a consequence of the definition, and so doesn't need to be included in the definition.
To answer your third question, "why is it clear from the formal definition that $x_n$ actually converges to $x$ as $n$ increases?", there's a temptation to ask "what do you mean by 'actually converges'?" Perhaps a helpful observation is that the definition is equivalent to saying: for any $epsilon > 0$, all but finitely many of the terms of the sequence are within $epsilon$ of $x$. We don't care how many that "finitely many" is, provided there are only finitely many of them.
[1]: I hope it's clear what I mean by "$N$ works for $epsilon$" here and later, I just don't want to have to keep writing out the second half of the definition
answered 50 mins ago
Christopher
5,44911526
add a comment |Â
up vote
2
down vote
You are right that $N$ depends on $epsilon$ - and missing this fact is the cause of a lot of confusion among people learning analysis for the first time. Sometimes it is made explicit: "for all $epsilon > 0$ there exists an $N$ (dependent on $epsilon$) such that...". But even when phrased as "for all $epsilon > 0$ there exists an $N$ such that...", $N$ still depends on $epsilon$: that's how quantifiers ("for all", "there exists") work, even if it's not made explicit.
One thing that should be pointed out is that $N$ is not unique. Indeed if, for a given $epsilon$, $N = M$ works[1], then $N = M'$ works for any $M' geq M$. We could define things such that $N$ was the minimum that worked, but this would make both the definition and proofs involving the definition more convoluted, with very little gain in return.
With this in mind, it's somewhat unhelpful to think of $N$ as a function of $epsilon$, because "function" suggests there's a unique value of $N$ for each $epsilon$, which isn't true.
On the topic of $N$ increasing as $epsilon$ decreases: this is not true in the most pedantic sense: if $x_i = 0$ for all $i$, then we could take $N = 10$ when $epsilon > 1$ and $N = 1$ when $epsilon leq 1$, the definition is satisfied but $N$ is not increasing as $epsilon$ decreases.
However, in a less pedantic sense, it is true that if $N$ works for $epsilon$ and $epsilon' > epsilon$ then $N$ works for $epsilon'$ too. But this is a consequence of the definition, and so doesn't need to be included in the definition.
To answer your third question, "why is it clear from the formal definition that $x_n$ actually converges to $x$ as $n$ increases?", there's a temptation to ask "what do you mean by 'actually converges'?" Perhaps a helpful observation is that the definition is equivalent to saying: for any $epsilon > 0$, all but finitely many of the terms of the sequence are within $epsilon$ of $x$. We don't care how many that "finitely many" is, provided there are only finitely many of them.
[1]: I hope it's clear what I mean by "$N$ works for $epsilon$" here and later, I just don't want to have to keep writing out the second half of the definition
answered 50 mins ago
Christopher
5,44911526
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You are right that $N$ depends on $epsilon$ - and missing this fact is the cause of a lot of confusion among people learning analysis for the first time. Sometimes it is made explicit: "for all $epsilon > 0$ there exists an $N$ (dependent on $epsilon$) such that...". But even when phrased as "for all $epsilon > 0$ there exists an $N$ such that...", $N$ still depends on $epsilon$: that's how quantifiers ("for all", "there exists") work, even if it's not made explicit.
One thing that should be pointed out is that $N$ is not unique. Indeed if, for a given $epsilon$, $N = M$ works[1], then $N = M'$ works for any $M' geq M$. We could define things such that $N$ was the minimum that worked, but this would make both the definition and proofs involving the definition more convoluted, with very little gain in return.
With this in mind, it's somewhat unhelpful to think of $N$ as a function of $epsilon$, because "function" suggests there's a unique value of $N$ for each $epsilon$, which isn't true.
On the topic of $N$ increasing as $epsilon$ decreases: this is not true in the most pedantic sense: if $x_i = 0$ for all $i$, then we could take $N = 10$ when $epsilon > 1$ and $N = 1$ when $epsilon leq 1$, the definition is satisfied but $N$ is not increasing as $epsilon$ decreases.
However, in a less pedantic sense, it is true that if $N$ works for $epsilon$ and $epsilon' > epsilon$ then $N$ works for $epsilon'$ too. But this is a consequence of the definition, and so doesn't need to be included in the definition.
To answer your third question, "why is it clear from the formal definition that $x_n$ actually converges to $x$ as $n$ increases?", there's a temptation to ask "what do you mean by 'actually converges'?" Perhaps a helpful observation is that the definition is equivalent to saying: for any $epsilon > 0$, all but finitely many of the terms of the sequence are within $epsilon$ of $x$. We don't care how many that "finitely many" is, provided there are only finitely many of them.
[1]: I hope it's clear what I mean by "$N$ works for $epsilon$" here and later, I just don't want to have to keep writing out the second half of the definition
answered 50 mins ago
Christopher
5,44911526
You are right that $N$ depends on $epsilon$ - and missing this fact is the cause of a lot of confusion among people learning analysis for the first time. Sometimes it is made explicit: "for all $epsilon > 0$ there exists an $N$ (dependent on $epsilon$) such that...". But even when phrased as "for all $epsilon > 0$ there exists an $N$ such that...", $N$ still depends on $epsilon$: that's how quantifiers ("for all", "there exists") work, even if it's not made explicit.
One thing that should be pointed out is that $N$ is not unique. Indeed if, for a given $epsilon$, $N = M$ works[1], then $N = M'$ works for any $M' geq M$. We could define things such that $N$ was the minimum that worked, but this would make both the definition and proofs involving the definition more convoluted, with very little gain in return.
With this in mind, it's somewhat unhelpful to think of $N$ as a function of $epsilon$, because "function" suggests there's a unique value of $N$ for each $epsilon$, which isn't true.
On the topic of $N$ increasing as $epsilon$ decreases: this is not true in the most pedantic sense: if $x_i = 0$ for all $i$, then we could take $N = 10$ when $epsilon > 1$ and $N = 1$ when $epsilon leq 1$, the definition is satisfied but $N$ is not increasing as $epsilon$ decreases.
However, in a less pedantic sense, it is true that if $N$ works for $epsilon$ and $epsilon' > epsilon$ then $N$ works for $epsilon'$ too. But this is a consequence of the definition, and so doesn't need to be included in the definition.
To answer your third question, "why is it clear from the formal definition that $x_n$ actually converges to $x$ as $n$ increases?", there's a temptation to ask "what do you mean by 'actually converges'?" Perhaps a helpful observation is that the definition is equivalent to saying: for any $epsilon > 0$, all but finitely many of the terms of the sequence are within $epsilon$ of $x$. We don't care how many that "finitely many" is, provided there are only finitely many of them.
[1]: I hope it's clear what I mean by "$N$ works for $epsilon$" here and later, I just don't want to have to keep writing out the second half of the definition
answered 50 mins ago
Christopher
5,44911526
answered 50 mins ago
Christopher
5,44911526
answered 50 mins ago
Christopher
5,44911526
answered 50 mins ago
Christopher
5,44911526
5,44911526
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2956460%2fshould-the-formal-definition-of-the-limit-of-a-sequence-be-improved%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
$N$ is never unique.
– Randall
1 hour ago
Also consider constant sequences.
– Randall
1 hour ago
1
Perhaps you are thinking of $N$ as the minimal integer such that $|x_n - x|< epsilon$ for $n ge N$. But that is not a practical definition.
– Martin R
1 hour ago
$N$ would also depend on the sequence, so $N(epsilon, a_1,a_2,ldots)$? Seriously, we do not use functional notation in this place, the correct nesting of quantors is enough (and works with the common rules of inference involving them)
– Hagen von Eitzen
1 hour ago
@MartinR, this is indeed what I am at. Why is it not a practical definition? This is what the figure of the article says right? As $N$ increases, $varepsilon$ must decrease, right? Otherwise convergence does not take place.
– Snoopy
1 hour ago