Proof verification. Show that if ${x_n}$ diverges then there must be a sequence ${p_n}subsetBbb N$ such that...
$begingroup$
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
asked Dec 26 '18 at 15:53
romanroman
2,50721226
$endgroup$
add a comment |
$begingroup$
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
asked Dec 26 '18 at 15:53
romanroman
2,50721226
$endgroup$
1
$begingroup$
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
$endgroup$
– Michał Miśkiewicz
Dec 26 '18 at 18:03
$begingroup$
A visualization of the idea from @RafayAshary answer. Just for the case
$endgroup$
– roman
Dec 26 '18 at 18:36
add a comment |
$begingroup$
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
asked Dec 26 '18 at 15:53
romanroman
2,50721226
$endgroup$
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
calculus limits proof-verification convergence
asked Dec 26 '18 at 15:53
romanroman
2,50721226
asked Dec 26 '18 at 15:53
romanroman
2,50721226
edited Dec 26 '18 at 16:11
roman
asked Dec 26 '18 at 15:53
romanroman
2,50721226
asked Dec 26 '18 at 15:53
romanroman
2,50721226
asked Dec 26 '18 at 15:53
romanroman
2,50721226
2,50721226
1
$begingroup$
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
$endgroup$
– Michał Miśkiewicz
Dec 26 '18 at 18:03
$begingroup$
A visualization of the idea from @RafayAshary answer. Just for the case
$endgroup$
– roman
Dec 26 '18 at 18:36
add a comment |
1
$begingroup$
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
$endgroup$
– Michał Miśkiewicz
Dec 26 '18 at 18:03
$begingroup$
A visualization of the idea from @RafayAshary answer. Just for the case
$endgroup$
– roman
Dec 26 '18 at 18:36
1
1
$begingroup$
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
$endgroup$
– Michał Miśkiewicz
Dec 26 '18 at 18:03
$begingroup$
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
$endgroup$
– Michał Miśkiewicz
Dec 26 '18 at 18:03
$begingroup$
A visualization of the idea from @RafayAshary answer. Just for the case
$endgroup$
– roman
Dec 26 '18 at 18:36
$begingroup$
A visualization of the idea from @RafayAshary answer. Just for the case
$endgroup$
– roman
Dec 26 '18 at 18:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
$endgroup$
2
$begingroup$
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
$endgroup$
– John Doe
Dec 26 '18 at 18:01
$begingroup$
Yes, my apologies :)
$endgroup$
– Rafay Ashary
Dec 26 '18 at 18:19
$begingroup$
That's a nice way to approach the proof, thank you!
$endgroup$
– roman
Dec 26 '18 at 18:37
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053057%2fproof-verification-show-that-if-x-n-diverges-then-there-must-be-a-sequenc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
$endgroup$
2
$begingroup$
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
$endgroup$
– John Doe
Dec 26 '18 at 18:01
$begingroup$
Yes, my apologies :)
$endgroup$
– Rafay Ashary
Dec 26 '18 at 18:19
$begingroup$
That's a nice way to approach the proof, thank you!
$endgroup$
– roman
Dec 26 '18 at 18:37
add a comment |
$begingroup$
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
$endgroup$
2
$begingroup$
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
$endgroup$
– John Doe
Dec 26 '18 at 18:01
$begingroup$
Yes, my apologies :)
$endgroup$
– Rafay Ashary
Dec 26 '18 at 18:19
$begingroup$
That's a nice way to approach the proof, thank you!
$endgroup$
– roman
Dec 26 '18 at 18:37
add a comment |
$begingroup$
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
$endgroup$
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
edited Dec 26 '18 at 18:19
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
answered Dec 26 '18 at 17:56
Rafay AsharyRafay Ashary
84618
84618
2
$begingroup$
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
$endgroup$
– John Doe
Dec 26 '18 at 18:01
$begingroup$
Yes, my apologies :)
$endgroup$
– Rafay Ashary
Dec 26 '18 at 18:19
$begingroup$
That's a nice way to approach the proof, thank you!
$endgroup$
– roman
Dec 26 '18 at 18:37
add a comment |
2
$begingroup$
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
$endgroup$
– John Doe
Dec 26 '18 at 18:01
$begingroup$
Yes, my apologies :)
$endgroup$
– Rafay Ashary
Dec 26 '18 at 18:19
$begingroup$
That's a nice way to approach the proof, thank you!
$endgroup$
– roman
Dec 26 '18 at 18:37
2
2
$begingroup$
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
$endgroup$
– John Doe
Dec 26 '18 at 18:01
$begingroup$
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
$endgroup$
– John Doe
Dec 26 '18 at 18:01
$begingroup$
Yes, my apologies :)
$endgroup$
– Rafay Ashary
Dec 26 '18 at 18:19
$begingroup$
Yes, my apologies :)
$endgroup$
– Rafay Ashary
Dec 26 '18 at 18:19
$begingroup$
That's a nice way to approach the proof, thank you!
$endgroup$
– roman
Dec 26 '18 at 18:37
$begingroup$
That's a nice way to approach the proof, thank you!
$endgroup$
– roman
Dec 26 '18 at 18:37
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053057%2fproof-verification-show-that-if-x-n-diverges-then-there-must-be-a-sequenc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
$endgroup$
– Michał Miśkiewicz
Dec 26 '18 at 18:03
$begingroup$
A visualization of the idea from @RafayAshary answer. Just for the case
$endgroup$
– roman
Dec 26 '18 at 18:36