Show that the following sequence converges. Please Critique my proof.Prove that a sequence converges to a...
Can you earn endless XP using a Flameskull and its self-revival feature?
Why doesn't "auto ch = unsigned char{'p'}" compile under C++ 17?
How to deal with an incendiary email that was recalled
How to apply float precision (type specifier) in a conditional f-string?
Book where aliens are selecting humans for food consumption
Why does a metal block make a shrill sound but not a wooden block upon hammering?
Contest math problem about crossing out numbers in the table
Would these multi-classing house rules cause unintended problems?
Dilemma of explaining to interviewer that he is the reason for declining second interview
Program that converts a number to a letter of the alphabet
Word or phrase for showing great skill at something without formal training in it
Can I become debt free or should I file for bankruptcy? How do I manage my debt and finances?
Explain the objections to these measures against human trafficking
Typing Amharic inside a math equation?
If I sold a PS4 game I owned the disc for, can I reinstall it digitally?
Does Improved Divine Strike trigger when a paladin makes an unarmed strike?
Why do members of Congress in committee hearings ask witnesses the same question multiple times?
A minimum of two personnel "are" or "is"?
Broken patches on a road
Are there neural networks with very few nodes that decently solve non-trivial problems?
How can animals be objects of ethics without being subjects as well?
Can a hotel cancel a confirmed reservation?
Avoiding morning and evening handshakes
figures in a grid with multiple line of texts
Show that the following sequence converges. Please Critique my proof.
Prove that a sequence converges to a finite limit iff lim inf equals lim supShow the convergence of sequenceSimple proof that this sequence converges [verification]Proof that the sequence $a_n=frac{3n+2}{n^2+1}$ converges using the Epsilon N proofQuestion about the proof that the sequence ${a_{j}cdot b_{j}}$ converges to $alpha beta $Show that the sequence ${a_{n+1}}$ converges to $sqrt{2}$Showing that a series converges uniformlyshow that if a subsequence of a cauchy sequence converges, then the whole sequence convergesShow that the sequence $a_n=frac{cos(n^2+n)}{n^2}$ converges to $0$.Proof that the sequence $left{frac{5n^2-6}{2n^3-7n}right}$ converges to $0$
$begingroup$
The problem is as follows:
Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.
My (wrong) proof:
Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.
My question:
This is a question from a comprehensive exam I found and am using to review.
Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?
Notes: Currently working on the proof.
real-analysis sequences-and-series convergence fake-proofs
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
asked 2 hours ago
DarelDarel
1049
$endgroup$
add a comment |
$begingroup$
The problem is as follows:
Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.
My (wrong) proof:
Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.
My question:
This is a question from a comprehensive exam I found and am using to review.
Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?
Notes: Currently working on the proof.
real-analysis sequences-and-series convergence fake-proofs
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
asked 2 hours ago
DarelDarel
1049
$endgroup$
4
$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago
$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
34 mins ago
add a comment |
$begingroup$
The problem is as follows:
Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.
My (wrong) proof:
Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.
My question:
This is a question from a comprehensive exam I found and am using to review.
Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?
Notes: Currently working on the proof.
real-analysis sequences-and-series convergence fake-proofs
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
asked 2 hours ago
DarelDarel
1049
$endgroup$
The problem is as follows:
Let ${a_n}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}leq a_n+frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.
My (wrong) proof:
Notice that
$$
|a_{n+1}-a_n|leq left|frac{(-1)^n}{n}right|leqfrac{1}{n}
$$
and since it is known that $frac{1}{n}rightarrow 0$ as $nrightarrow infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.
My question:
This is a question from a comprehensive exam I found and am using to review.
Should I argue that we should select $N$ so that $n>N$ implies $left|frac{1}{n}right|<epsilon$ as well?
Notes: Currently working on the proof.
real-analysis sequences-and-series convergence fake-proofs
real-analysis sequences-and-series convergence fake-proofs
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
asked 2 hours ago
DarelDarel
1049
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
asked 2 hours ago
DarelDarel
1049
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
edited 32 mins ago
GNUSupporter 8964民主女神 地下教會
13.9k72650
13.9k72650
asked 2 hours ago
DarelDarel
1049
asked 2 hours ago
DarelDarel
1049
asked 2 hours ago
DarelDarel
1049
1049
4
$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago
$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
34 mins ago
add a comment |
4
$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago
$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
34 mins ago
4
4
$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago
$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago
$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
34 mins ago
$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
34 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then
$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$
which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
2
$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
19 mins ago
add a comment |
$begingroup$
As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$
EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
$endgroup$
$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago
$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
23 mins ago
$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
18 mins ago
$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
25 secs ago
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
28 mins ago
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%2f3131816%2fshow-that-the-following-sequence-converges-please-critique-my-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then
$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$
which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
2
$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
19 mins ago
add a comment |
$begingroup$
Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then
$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$
which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
2
$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
19 mins ago
add a comment |
$begingroup$
Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then
$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$
which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
$endgroup$
Consider $b_n = a_n + sum_{k=1}^{n-1} frac{(-1)^{k-1}}{k}$. Then
$$ b_{n+1}
= a_{n+1} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
leq a_n + frac{(-1)^n}{n} + sum_{k=1}^{n} frac{(-1)^{k-1}}{k}
= b_n, $$
which shows that $(b_n)$ is non-increasing. Moreover, since $sum_{k=1}^{infty} frac{(-1)^{k-1}}{k}$ converges by alternating series test and $(a_n)$ is non-negative, it follows that $(b_n)$ is bounded from below. Therefore $(b_n)$ converges, and so, $(a_n)$ converges as well.
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
answered 36 mins ago
Sangchul LeeSangchul Lee
95k12170276
95k12170276
2
$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
19 mins ago
add a comment |
2
$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
19 mins ago
2
2
$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
19 mins ago
$begingroup$
Thank you, that's neat! One might add that this argument always works for lower-bounded $(a_n)$ with $a_{n+1}le a_n+c_n$ for some summable $(c_n)$ by setting $b_n=a_n-sum_{k=1}^{n-1}c_k$.
$endgroup$
– Mars Plastic
19 mins ago
add a comment |
$begingroup$
As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$
EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
$endgroup$
$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago
$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
23 mins ago
$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
18 mins ago
$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
25 secs ago
add a comment |
$begingroup$
As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$
EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
$endgroup$
$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago
$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
23 mins ago
$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
18 mins ago
$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
25 secs ago
add a comment |
$begingroup$
As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$
EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
$endgroup$
As pointed out in the comments, this approach does not work. However, you can show that $(a_n)$ is a Cauchy sequence by using that
$$ sum_{n=1}^infty (a_{n+1}-a_n)lesum_{n=1}^infty (-1)^nfrac{1}{n}<infty.$$
EDIT: I am sorry, that was a bit too fast. Since the sequence is not assumed to be increasing, this alone does not help. I wanted to delete the answer, but it has already been accepted which apparently makes that impossible.
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
edited 50 mins ago
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
answered 2 hours ago
Mars PlasticMars Plastic
1,07918
1,07918
$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago
$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
23 mins ago
$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
18 mins ago
$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
25 secs ago
add a comment |
$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago
$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
23 mins ago
$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
18 mins ago
$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
25 secs ago
$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago
$begingroup$
Okay kind of a dumb question here. I forgot how this would imply that the sequence is Cauchy. Do you have a source where I can see this for myself? I know I saw it somewhere but my memory isn't working for me. I think one of the key thing was that all bounded monotone sequences converge or something like that.
$endgroup$
– Darel
1 hour ago
$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
23 mins ago
$begingroup$
That doesn't show that it's Cauchy. Cauchy doesn't mean that the difference between successive terms goes to zero, it mean that $max_{n>N} |a_N-a_n|$ goes to zero as N goes to zero.
$endgroup$
– Acccumulation
23 mins ago
$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
18 mins ago
$begingroup$
1.) I did not claim that, not even in my original answer. 2.) Read my edit.
$endgroup$
– Mars Plastic
18 mins ago
$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
25 secs ago
$begingroup$
I unlocked it and gave appropriate credit to the clearer answer.
$endgroup$
– Darel
25 secs ago
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
28 mins ago
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
28 mins ago
add a comment |
$begingroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
edited 1 hour ago
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
answered 1 hour ago
Michael RozenbergMichael Rozenberg
106k1893198
106k1893198
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
28 mins ago
add a comment |
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
28 mins ago
1
1
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago
$begingroup$
Be careful! The assumption is only an inequality.
$endgroup$
– Mars Plastic
1 hour ago
1
1
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
@Mars Plastic I see. It was typo.
$endgroup$
– Michael Rozenberg
1 hour ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
28 mins ago
$begingroup$
That shows that $(a_n)$ is bounded above, but why is it convergent?
$endgroup$
– Martin R
28 mins ago
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%2f3131816%2fshow-that-the-following-sequence-converges-please-critique-my-proof%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
4
$begingroup$
Your proof is not correct. Your arguments would also work for $a_n = sum_{i=1}^n frac 1 i$, which does not converge.
$endgroup$
– Falrach
2 hours ago
$begingroup$
Note that you not only need to bound $left| a_{n+1} - a_n right|$ arbitrarily small, but also $left| a_{m} - a_n right|$ for all $m,n geq N$ (where $N$ can be chosen according to the bound).
$endgroup$
– Maximilian Janisch
34 mins ago