Dtjytyk

Why would the IRS ask for birth certificates or even audit a small tax return?

New invention compresses matter to produce energy? or other items? (Short Story)

The past tense for the quoting particle って

Short story about an infectious indestructible metal bar?

Create chunks from an array

Is divide-by-zero a security vulnerability?

What is the purpose of a disclaimer like "this is not legal advice"?

Why is my explanation wrong?

In the world of The Matrix, what is "popping"?

What is the meaning of option 'by' in TikZ Intersections

Ignoring Someone as Wrongful Speech

What's the best tool for cutting holes into duct work?

If nine coins are tossed, what is the probability that the number of heads is even?

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

Paper published similar to PhD thesis

Using the imperfect indicative vs. subjunctive with si

Python 3.6+ function to ask for a multiple-choice answer

Why do phishing e-mails use faked e-mail addresses instead of the real one?

Why aren't there more gauls like Obelix?

Can a Mimic (container form) actually hold loot?

Professor forcing me to attend a conference

Infinitive vs Gerund

Should we avoid writing fiction about historical events without extensive research?

How can I be pwned if I'm not registered on the compromised site?

Proving that a sequence converges if $|a_n - a_{n+1}| 0$ and $r in (0,1).$

Proof verification. ${x_n}$ is a sequence such that $|x_{n+1} - x_n| le Calpha^n$ for $alphain (0, 1), ninBbb N$. Prove $x_n$ converges.Proving a Sequence is CauchyProving that the sequence convergesProving that if $a_ngeq0$ and $sum a_n$ converges, then $sum a_n^2$ convergesIf $sum n a_n$ converges, then $sum a_n$ converges and $sum |a_n|^p$ converges for $p>1$.Prove that the following sequence converges to zero.Proving that a sequence $|a_n|leq 1/n$ is Cauchy.Proving a Sequence Converges - Cauchy?Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is CauchyProving convergence of a sequence with logarithmLet there be a sequence such that the distance between two consecutive terms converges to 0. Must this sequence converge?

2

1

$begingroup$

Let ${a_n}_n$ be a sequence. Suppose that there exist $M gt 0$ and $r in (0, 1)$ such that $|a_n - a_{n+1}| lt Mr^n$ for all $ n in Bbb N.$ Prove that ${a_n}_n$ converges. I'm not really sure where to go about with this problem. I feel like the easiest way would be to prove that the sequence is Cauchy, but I'm not entirely sure. Any help would be appreciated, thanks!

real-analysis sequences-and-series cauchy-sequences

share|cite|improve this question

edited 17 hours ago

mechanodroid

28.4k62548

asked 17 hours ago

B RetnikB Retnik

534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer I've added is based on a very similar question I've also asked here some time ago. I've only adapted it to use you notation. You may still want to take a look at the original question.
  $endgroup$
  – roman
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of the link from roman’s comment.
  $endgroup$
  – Shalop
  16 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Let ${a_n}_n$ be a sequence. Suppose that there exist $M gt 0$ and $r in (0, 1)$ such that $|a_n - a_{n+1}| lt Mr^n$ for all $ n in Bbb N.$ Prove that ${a_n}_n$ converges. I'm not really sure where to go about with this problem. I feel like the easiest way would be to prove that the sequence is Cauchy, but I'm not entirely sure. Any help would be appreciated, thanks!

real-analysis sequences-and-series cauchy-sequences

share|cite|improve this question

edited 17 hours ago

mechanodroid

28.4k62548

asked 17 hours ago

B RetnikB Retnik

534

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The answer I've added is based on a very similar question I've also asked here some time ago. I've only adapted it to use you notation. You may still want to take a look at the original question.
  $endgroup$
  – roman
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of the link from roman’s comment.
  $endgroup$
  – Shalop
  16 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let ${a_n}_n$ be a sequence. Suppose that there exist $M gt 0$ and $r in (0, 1)$ such that $|a_n - a_{n+1}| lt Mr^n$ for all $ n in Bbb N.$ Prove that ${a_n}_n$ converges. I'm not really sure where to go about with this problem. I feel like the easiest way would be to prove that the sequence is Cauchy, but I'm not entirely sure. Any help would be appreciated, thanks!

real-analysis sequences-and-series cauchy-sequences

share|cite|improve this question

edited 17 hours ago

mechanodroid

28.4k62548

asked 17 hours ago

B RetnikB Retnik

534

$endgroup$

share|cite|improve this question

edited 17 hours ago

mechanodroid

28.4k62548

asked 17 hours ago

B RetnikB Retnik

534

share|cite|improve this question

edited 17 hours ago

mechanodroid

28.4k62548

asked 17 hours ago

B RetnikB Retnik

534

edited 17 hours ago

mechanodroid

28.4k62548

edited 17 hours ago

mechanodroid

28.4k62548

asked 17 hours ago

B RetnikB Retnik

534

asked 17 hours ago

B RetnikB Retnik

534

3 Answers
3

active

oldest

votes

0

$begingroup$

You may start by considering the following set of inequalities:
$$
begin{align}
|x_{n+1} - x_n| &< Mr^n\
|x_{n+2} - x_{n+1}| &< Mr^{n+1}\
|x_{n+3} - x_{n+2}| &< Mr^{n+2}\
&cdots \
|x_{n+p+1} - x_{n+p}| &< Mr^{n+p}
end{align}
$$

Sum up the inequalities to obtain:
$$
|x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}| < Msum_{k=0}^pr^{n+k}
$$

By geometric sum:
$$
Msum_{k=0}^pr^{n+k} = Mfrac{r^n(1-r^{p+1})}{1-r} le Mfrac{r^n}{1-r}
$$

You can now choose some $epsilon > 0$ and $N in Bbb N$ such that:
$$
Mfrac{r^N}{1-r} < epsilon tag1
$$

Define:
$$
r = frac{1}{1+q}, qinBbb R_{>0}
$$

Thus:
$$
Mfrac{r^N}{1-r} = Mfrac{1}{(1-r)(1+q)^N} < epsilon
$$

Or:
$$
N > log_{1+q}frac{M}{(1-r)epsilon} implies Mfrac{r^N}{1-r} < epsilon
$$

Now going back to the sum of absolute values and using triangular inequality one may obtain:
$$
|x_{n+1} - x_n + x_{n+2} - x_{n+1} + cdots + x_{n+p+1} - x_{n+p}| le |x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}|
$$

Note the telescoping nature and you get:
$$
|x_{n+1} - x_{n+p+1}| < Mfrac{r^n}{1-r} < epsilon
$$

Now for all $n$ such that:
$$
n > N > log_{1+q}frac{M}{(1-r)epsilon}
$$

the inequality $(1)$ holds, which shows that the sequence satisfies Cauchy's criteria hence convergent.

share|cite|improve this answer

answered 16 hours ago

romanroman

2,34321224

$endgroup$

add a comment | 

1

$begingroup$

Hint: for $m > n$,
$$|a_n - a_m|le M(r^n + r^{n+1} +cdots + r^{m-1})$$
(sum of a geometric progression $=cdots$ ?)

share|cite|improve this answer

answered 17 hours ago

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.5k42971

$endgroup$

add a comment | 

1

$begingroup$

We have
$$sum_{n=1}^infty |a_{n+1}-a_n| le sum_{n=1}^infty Mr^n = frac{Mr}{1-r} < +infty$$
Absolute convergence of a series implies convergence so
$$a_1 + sum_{n=1}^infty(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} sum_{n=1}^{N-1}(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} (a_N - a_1) = lim_{Ntoinfty} a_N$$
also exists.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

mechanodroidmechanodroid

28.4k62548

$endgroup$

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%2f3138871%2fproving-that-a-sequence-converges-if-a-n-a-n1-mrn-for-some-m-0-a%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

You may start by considering the following set of inequalities:
$$
begin{align}
|x_{n+1} - x_n| &< Mr^n\
|x_{n+2} - x_{n+1}| &< Mr^{n+1}\
|x_{n+3} - x_{n+2}| &< Mr^{n+2}\
&cdots \
|x_{n+p+1} - x_{n+p}| &< Mr^{n+p}
end{align}
$$

Sum up the inequalities to obtain:
$$
|x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}| < Msum_{k=0}^pr^{n+k}
$$

By geometric sum:
$$
Msum_{k=0}^pr^{n+k} = Mfrac{r^n(1-r^{p+1})}{1-r} le Mfrac{r^n}{1-r}
$$

You can now choose some $epsilon > 0$ and $N in Bbb N$ such that:
$$
Mfrac{r^N}{1-r} < epsilon tag1
$$

Define:
$$
r = frac{1}{1+q}, qinBbb R_{>0}
$$

Thus:
$$
Mfrac{r^N}{1-r} = Mfrac{1}{(1-r)(1+q)^N} < epsilon
$$

Or:
$$
N > log_{1+q}frac{M}{(1-r)epsilon} implies Mfrac{r^N}{1-r} < epsilon
$$

Now going back to the sum of absolute values and using triangular inequality one may obtain:
$$
|x_{n+1} - x_n + x_{n+2} - x_{n+1} + cdots + x_{n+p+1} - x_{n+p}| le |x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}|
$$

Note the telescoping nature and you get:
$$
|x_{n+1} - x_{n+p+1}| < Mfrac{r^n}{1-r} < epsilon
$$

Now for all $n$ such that:
$$
n > N > log_{1+q}frac{M}{(1-r)epsilon}
$$

the inequality $(1)$ holds, which shows that the sequence satisfies Cauchy's criteria hence convergent.

share|cite|improve this answer

answered 16 hours ago

romanroman

2,34321224

$endgroup$

add a comment | 

0

$begingroup$

You may start by considering the following set of inequalities:
$$
begin{align}
|x_{n+1} - x_n| &< Mr^n\
|x_{n+2} - x_{n+1}| &< Mr^{n+1}\
|x_{n+3} - x_{n+2}| &< Mr^{n+2}\
&cdots \
|x_{n+p+1} - x_{n+p}| &< Mr^{n+p}
end{align}
$$

Sum up the inequalities to obtain:
$$
|x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}| < Msum_{k=0}^pr^{n+k}
$$

By geometric sum:
$$
Msum_{k=0}^pr^{n+k} = Mfrac{r^n(1-r^{p+1})}{1-r} le Mfrac{r^n}{1-r}
$$

You can now choose some $epsilon > 0$ and $N in Bbb N$ such that:
$$
Mfrac{r^N}{1-r} < epsilon tag1
$$

Define:
$$
r = frac{1}{1+q}, qinBbb R_{>0}
$$

Thus:
$$
Mfrac{r^N}{1-r} = Mfrac{1}{(1-r)(1+q)^N} < epsilon
$$

Or:
$$
N > log_{1+q}frac{M}{(1-r)epsilon} implies Mfrac{r^N}{1-r} < epsilon
$$

Now going back to the sum of absolute values and using triangular inequality one may obtain:
$$
|x_{n+1} - x_n + x_{n+2} - x_{n+1} + cdots + x_{n+p+1} - x_{n+p}| le |x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}|
$$

Note the telescoping nature and you get:
$$
|x_{n+1} - x_{n+p+1}| < Mfrac{r^n}{1-r} < epsilon
$$

Now for all $n$ such that:
$$
n > N > log_{1+q}frac{M}{(1-r)epsilon}
$$

the inequality $(1)$ holds, which shows that the sequence satisfies Cauchy's criteria hence convergent.

share|cite|improve this answer

answered 16 hours ago

romanroman

2,34321224

$endgroup$

add a comment | 

$begingroup$

You may start by considering the following set of inequalities:
$$
begin{align}
|x_{n+1} - x_n| &< Mr^n\
|x_{n+2} - x_{n+1}| &< Mr^{n+1}\
|x_{n+3} - x_{n+2}| &< Mr^{n+2}\
&cdots \
|x_{n+p+1} - x_{n+p}| &< Mr^{n+p}
end{align}
$$

Sum up the inequalities to obtain:
$$
|x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}| < Msum_{k=0}^pr^{n+k}
$$

By geometric sum:
$$
Msum_{k=0}^pr^{n+k} = Mfrac{r^n(1-r^{p+1})}{1-r} le Mfrac{r^n}{1-r}
$$

You can now choose some $epsilon > 0$ and $N in Bbb N$ such that:
$$
Mfrac{r^N}{1-r} < epsilon tag1
$$

Define:
$$
r = frac{1}{1+q}, qinBbb R_{>0}
$$

Thus:
$$
Mfrac{r^N}{1-r} = Mfrac{1}{(1-r)(1+q)^N} < epsilon
$$

Or:
$$
N > log_{1+q}frac{M}{(1-r)epsilon} implies Mfrac{r^N}{1-r} < epsilon
$$

Now going back to the sum of absolute values and using triangular inequality one may obtain:
$$
|x_{n+1} - x_n + x_{n+2} - x_{n+1} + cdots + x_{n+p+1} - x_{n+p}| le |x_{n+1} - x_n| + |x_{n+2} - x_{n+1}| + cdots + |x_{n+p+1} - x_{n+p}|
$$

Note the telescoping nature and you get:
$$
|x_{n+1} - x_{n+p+1}| < Mfrac{r^n}{1-r} < epsilon
$$

Now for all $n$ such that:
$$
n > N > log_{1+q}frac{M}{(1-r)epsilon}
$$

the inequality $(1)$ holds, which shows that the sequence satisfies Cauchy's criteria hence convergent.

share|cite|improve this answer

answered 16 hours ago

romanroman

2,34321224

$endgroup$

share|cite|improve this answer

answered 16 hours ago

romanroman

2,34321224

answered 16 hours ago

romanroman

2,34321224

answered 16 hours ago

romanroman

2,34321224

1

$begingroup$

Hint: for $m > n$,
$$|a_n - a_m|le M(r^n + r^{n+1} +cdots + r^{m-1})$$
(sum of a geometric progression $=cdots$ ?)

share|cite|improve this answer

answered 17 hours ago

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.5k42971

$endgroup$

add a comment | 

1

$begingroup$

Hint: for $m > n$,
$$|a_n - a_m|le M(r^n + r^{n+1} +cdots + r^{m-1})$$
(sum of a geometric progression $=cdots$ ?)

share|cite|improve this answer

answered 17 hours ago

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.5k42971

$endgroup$

add a comment | 

$begingroup$

Hint: for $m > n$,
$$|a_n - a_m|le M(r^n + r^{n+1} +cdots + r^{m-1})$$
(sum of a geometric progression $=cdots$ ?)

share|cite|improve this answer

answered 17 hours ago

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.5k42971

$endgroup$

share|cite|improve this answer

answered 17 hours ago

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.5k42971

answered 17 hours ago

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.5k42971

answered 17 hours ago

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

34.5k42971

1

$begingroup$

We have
$$sum_{n=1}^infty |a_{n+1}-a_n| le sum_{n=1}^infty Mr^n = frac{Mr}{1-r} < +infty$$
Absolute convergence of a series implies convergence so
$$a_1 + sum_{n=1}^infty(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} sum_{n=1}^{N-1}(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} (a_N - a_1) = lim_{Ntoinfty} a_N$$
also exists.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

mechanodroidmechanodroid

28.4k62548

$endgroup$

add a comment | 

1

$begingroup$

We have
$$sum_{n=1}^infty |a_{n+1}-a_n| le sum_{n=1}^infty Mr^n = frac{Mr}{1-r} < +infty$$
Absolute convergence of a series implies convergence so
$$a_1 + sum_{n=1}^infty(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} sum_{n=1}^{N-1}(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} (a_N - a_1) = lim_{Ntoinfty} a_N$$
also exists.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

mechanodroidmechanodroid

28.4k62548

$endgroup$

add a comment | 

$begingroup$

We have
$$sum_{n=1}^infty |a_{n+1}-a_n| le sum_{n=1}^infty Mr^n = frac{Mr}{1-r} < +infty$$
Absolute convergence of a series implies convergence so
$$a_1 + sum_{n=1}^infty(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} sum_{n=1}^{N-1}(a_{n+1} - a_n) = a_1 + lim_{Ntoinfty} (a_N - a_1) = lim_{Ntoinfty} a_N$$
also exists.

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

mechanodroidmechanodroid

28.4k62548

$endgroup$

share|cite|improve this answer

edited 16 hours ago

answered 17 hours ago

mechanodroidmechanodroid

28.4k62548

answered 17 hours ago

mechanodroidmechanodroid

28.4k62548

answered 17 hours ago

mechanodroidmechanodroid

28.4k62548