Fdtxjr

What should you do if you miss a job interview (deliberately)?

How should I respond when I lied about my education and the company finds out through background check?

Creepy dinosaur pc game identification

Are paving bricks differently sized for sand bedding vs mortar bedding?

What does "Scientists rise up against statistical significance" mean? (Comment in Nature)

Electoral considerations aside, what are potential benefits, for the US, of policy changes proposed by the tweet recognizing Golan annexation?

Delivering sarcasm

Not using 's' for he/she/it

Count the occurrence of each unique word in the file

How do I color the graph in datavisualization?

Problem with TransformedDistribution

Is it possible to put a rectangle as background in the author section?

Which one is correct as adjective “protruding” or “protruded”?

How to implement a feedback to keep the DC gain at zero for this conceptual passive filter?

Can I sign legal documents with a smiley face?

What if a revenant (monster) gains fire resistance?

Is there any references on the tensor product of presentable (1-)categories?

If infinitesimal transformations commute why dont the generators of the Lorentz group commute?

Biological Blimps: Propulsion

Why did the Mercure fail?

Longest common substring in linear time

Drawing ramified coverings with tikz

Why electric field inside a cavity of a non-conducting sphere not zero?

2.8 Why are collections grayed out? How can I open them?

Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$. Does $x_n$ converge to $1$?

(edited) Let $(x_n_n)$ be a sequence of positive real numbers which has no convergent subsequence. Prove lim $x_n=+infty$Prove $x_n$ converges to $a$ iff every subsequence of $x_n$ also converges to $a$.A sequence converges if and only if every subsequence converges?If $lim_x to infty x_n = liminf_n to infty x_n = limsup_n to infty x_n= -infty$ does it exist a convergent subsequence of $x_n$?Given subsequences converge, prove that the sequence converges.If all proper subsequences converge to same limit then the sequence converges.Show that if all convergent subsequences of a bounded sequence converge to $l$, the sequence itself must also converge to $l$.Let $(a_n)_n in mathbb N$ be a sequence such that all non-trivial (*) subsequence converges, does $(a_n)$ converge?Convergence of sequence where every subsequence of specific type convergesProof verification. If $x_n$ is a monotone sequence and it has a convergent subsequence $x_n_k$, then $x_n$ is convergent to the same limit.

0

$begingroup$

Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$.

Does $x_n$ converge to $1$?

I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$:
$$
k = 2:x_pk = x_2, x_4, dots, x_2p \
k = 3:x_pk = x_3, x_4, dots, x_3p \
cdots\
k = n:x_pk = x_n, x_2n, dots, x_np \
$$

All of the sequences above converge to $1$, namely:
$$
lim_ptoinftyx_2p = 1\
lim_ptoinftyx_3p = 1\
cdots\
lim_ptoinftyx_np = 1\
$$

At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set:
$$
X = bigcuplimits_k=1^infty x_pk
$$

Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $x_n$ which would imply that:
$$
lim_ntoinftyx_n = 1
$$

I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete?
Thank you!

real-analysis sequences-and-series limits proof-verification

share|cite|improve this question

edited Mar 15 at 17:28

roman

asked Mar 15 at 17:09

romanroman

2,39321225

$endgroup$

add a comment | 

0

$begingroup$

Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$.

Does $x_n$ converge to $1$?

I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$:
$$
k = 2:x_pk = x_2, x_4, dots, x_2p \
k = 3:x_pk = x_3, x_4, dots, x_3p \
cdots\
k = n:x_pk = x_n, x_2n, dots, x_np \
$$

All of the sequences above converge to $1$, namely:
$$
lim_ptoinftyx_2p = 1\
lim_ptoinftyx_3p = 1\
cdots\
lim_ptoinftyx_np = 1\
$$

At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set:
$$
X = bigcuplimits_k=1^infty x_pk
$$

Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $x_n$ which would imply that:
$$
lim_ntoinftyx_n = 1
$$

I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete?
Thank you!

real-analysis sequences-and-series limits proof-verification

share|cite|improve this question

edited Mar 15 at 17:28

roman

asked Mar 15 at 17:09

romanroman

2,39321225

$endgroup$

add a comment | 

$begingroup$

Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$.

Does $x_n$ converge to $1$?

I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$:
$$
k = 2:x_pk = x_2, x_4, dots, x_2p \
k = 3:x_pk = x_3, x_4, dots, x_3p \
cdots\
k = n:x_pk = x_n, x_2n, dots, x_np \
$$

All of the sequences above converge to $1$, namely:
$$
lim_ptoinftyx_2p = 1\
lim_ptoinftyx_3p = 1\
cdots\
lim_ptoinftyx_np = 1\
$$

At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set:
$$
X = bigcuplimits_k=1^infty x_pk
$$

Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $x_n$ which would imply that:
$$
lim_ntoinftyx_n = 1
$$

I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete?
Thank you!

real-analysis sequences-and-series limits proof-verification

share|cite|improve this question

edited Mar 15 at 17:28

roman

asked Mar 15 at 17:09

romanroman

2,39321225

$endgroup$

share|cite|improve this question

edited Mar 15 at 17:28

roman

asked Mar 15 at 17:09

romanroman

2,39321225

share|cite|improve this question

edited Mar 15 at 17:28

roman

asked Mar 15 at 17:09

romanroman

2,39321225

asked Mar 15 at 17:09

romanroman

2,39321225

asked Mar 15 at 17:09

romanroman

2,39321225

1 Answer
1

active

oldest

votes

1

$begingroup$

Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.

share|cite|improve this answer

answered Mar 15 at 17:11

mathworker21mathworker21

9,4561928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
  $endgroup$
  – roman
  Mar 15 at 17:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't get the relevance
  $endgroup$
  – mathworker21
  Mar 15 at 17:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
  $endgroup$
  – roman
  Mar 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x_n$ is not convergent
  $endgroup$
  – mathworker21
  Mar 15 at 18:23
  

  
  
  
  

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%2f3149555%2flet-x-n-be-a-sequence-such-that-forall-k-1-in-bbb-n-its-subsequence%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.

share|cite|improve this answer

answered Mar 15 at 17:11

mathworker21mathworker21

9,4561928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
  $endgroup$
  – roman
  Mar 15 at 17:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't get the relevance
  $endgroup$
  – mathworker21
  Mar 15 at 17:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
  $endgroup$
  – roman
  Mar 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x_n$ is not convergent
  $endgroup$
  – mathworker21
  Mar 15 at 18:23
  

  
  
  
  

add a comment | 

1

$begingroup$

Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.

share|cite|improve this answer

answered Mar 15 at 17:11

mathworker21mathworker21

9,4561928

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
  $endgroup$
  – roman
  Mar 15 at 17:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't get the relevance
  $endgroup$
  – mathworker21
  Mar 15 at 17:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
  $endgroup$
  – roman
  Mar 15 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x_n$ is not convergent
  $endgroup$
  – mathworker21
  Mar 15 at 18:23
  

  
  
  
  

add a comment | 

$begingroup$

Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.

share|cite|improve this answer

answered Mar 15 at 17:11

mathworker21mathworker21

9,4561928

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:11

mathworker21mathworker21

9,4561928

answered Mar 15 at 17:11

mathworker21mathworker21

9,4561928

answered Mar 15 at 17:11

mathworker21mathworker21

9,4561928