Mrthfx

Placing an adverb between a verb and an object?

What is a jet (unit) shown in Windows 10 calculator?

Why is working on the same position for more than 15 years not a red flag?

Why does a metal block make a shrill sound but not a wooden block upon hammering?

Why Normality assumption in linear regression

Strange Sign on Lab Door

What is the wife of a henpecked husband called?

Using only 1s, make 29 with the minimum number of digits

Avoiding morning and evening handshakes

Does Windows 10's telemetry include sending *.doc files if Word crashed?

What was the earliest start time of a Catholic mass before 1957?

How to explain planetary rings pulsating?

What is better: yes / no radio, or simple checkbox?

Why zero tolerance on nudity in space?

How do you funnel food off a cutting board?

How to tag distinct options/entities without giving any an implicit priority or suggested order?

Enable Advanced Currency Management using CLI

It took me a lot of time to make this, pls like. (YouTube Comments #1)

figures in a grid with multiple line of texts

Would these multi-classing house rules cause unintended problems?

What makes the Forgotten Realms "forgotten"?

What does Cypher mean when he says Neo is "gonna pop"?

Broken patches on a road

Explain the objections to these measures against human trafficking

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 sequenceWhat is wrong with the following proof?Simple 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 if a subsequence of a cauchy sequence converges, then the whole sequence convergesProof that bounded growth of a sequence implies convergenceShow 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 $0$

6

1

$begingroup$

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

share|cite|improve this question

edited 30 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
  32 mins ago
  
  
  

  
  
  
  

add a comment | 

6

1

$begingroup$

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

share|cite|improve this question

edited 30 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
  32 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

share|cite|improve this question

edited 30 mins ago

GNUSupporter 8964民主女神 地下教會

13.9k72650

asked 2 hours ago

DarelDarel

1049

$endgroup$

share|cite|improve this question

edited 30 mins ago

GNUSupporter 8964民主女神 地下教會

13.9k72650

asked 2 hours ago

DarelDarel

1049

share|cite|improve this question

edited 30 mins ago

GNUSupporter 8964民主女神 地下教會

13.9k72650

asked 2 hours ago

DarelDarel

1049

edited 30 mins ago

GNUSupporter 8964民主女神 地下教會

13.9k72650

edited 30 mins ago

GNUSupporter 8964民主女神 地下教會

13.9k72650

asked 2 hours ago

DarelDarel

1049

asked 2 hours ago

DarelDarel

1049

3 Answers
3

active

oldest

votes

3

$begingroup$

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.

share|cite|improve this answer

edited 48 mins ago

answered 1 hour ago

Mars PlasticMars Plastic

1,09418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1.) I did not claim that, not even in my original answer. 2.) Read my edit.
  $endgroup$
  – Mars Plastic
  15 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 34 mins ago

Sangchul LeeSangchul Lee

95k12170276

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  17 mins ago
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$

share|cite|improve this answer

edited 1 hour ago

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
  25 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
});


}
});

draft saved

draft discarded

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

Name

Email

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

Name

Email

Required, but never shown

3

$begingroup$

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.

share|cite|improve this answer

edited 48 mins ago

answered 1 hour ago

Mars PlasticMars Plastic

1,09418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1.) I did not claim that, not even in my original answer. 2.) Read my edit.
  $endgroup$
  – Mars Plastic
  15 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

edited 48 mins ago

answered 1 hour ago

Mars PlasticMars Plastic

1,09418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  1.) I did not claim that, not even in my original answer. 2.) Read my edit.
  $endgroup$
  – Mars Plastic
  15 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.

share|cite|improve this answer

edited 48 mins ago

answered 1 hour ago

Mars PlasticMars Plastic

1,09418

$endgroup$

share|cite|improve this answer

edited 48 mins ago

answered 1 hour ago

Mars PlasticMars Plastic

1,09418

answered 1 hour ago

Mars PlasticMars Plastic

1,09418

answered 1 hour ago

Mars PlasticMars Plastic

1,09418

2

$begingroup$

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.

share|cite|improve this answer

answered 34 mins ago

Sangchul LeeSangchul Lee

95k12170276

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  17 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 34 mins ago

Sangchul LeeSangchul Lee

95k12170276

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  17 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.

share|cite|improve this answer

answered 34 mins ago

Sangchul LeeSangchul Lee

95k12170276

$endgroup$

share|cite|improve this answer

answered 34 mins ago

Sangchul LeeSangchul Lee

95k12170276

answered 34 mins ago

Sangchul LeeSangchul Lee

95k12170276

answered 34 mins ago

Sangchul LeeSangchul Lee

95k12170276

0

$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$

share|cite|improve this answer

edited 1 hour ago

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
  25 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$

share|cite|improve this answer

edited 1 hour ago

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
  25 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Use $$sum_{k=1}^n(a_{k+1}-a_k)=a_{n+1}-a_1leqsum_{k=1}^nfrac{(-1)^k}{k}rightarrow -ln2$$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Michael RozenbergMichael Rozenberg

106k1893198

$endgroup$

share|cite|improve this answer

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