A bounded sequence cannot be divergent. True or false












0












$begingroup$


"A bounded sequence cannot be divergent." Is this statement true?
As far as I know a bounded sequence can either be convergent or finitely oscillating, it cannot be divergent since it cannot diverge to infinity being a bounded sequence. Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?






real-analysis calculus sequences-and-series







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    "Divergent" means "not convergent".
    $endgroup$
    – Hans Engler
    Dec 4 '18 at 14:09






  • 3




    $begingroup$
    Recall that a sequence is said to be divergent if it is not convergent. Hence if it oscillates, it is considered to be divergent.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:10










  • $begingroup$
    @MisterRiemann I use another terminology, that is 1) convergent: $a_n to Lin mathbb{R}$, 2) divergent: $a_n toinfty$ or $a_n to-infty$ 3) Not convergent nor divergent otherwise.
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:15






  • 1




    $begingroup$
    That is a nonstandard terminology then. [1] [2]
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:31








  • 1




    $begingroup$
    @gimusi I've never seen such a convention. In any case, even the attached link classifies any sequence that does not converge is as a divergent sequence.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:42


















0












$begingroup$


"A bounded sequence cannot be divergent." Is this statement true?
As far as I know a bounded sequence can either be convergent or finitely oscillating, it cannot be divergent since it cannot diverge to infinity being a bounded sequence. Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?






real-analysis calculus sequences-and-series







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    "Divergent" means "not convergent".
    $endgroup$
    – Hans Engler
    Dec 4 '18 at 14:09






  • 3




    $begingroup$
    Recall that a sequence is said to be divergent if it is not convergent. Hence if it oscillates, it is considered to be divergent.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:10










  • $begingroup$
    @MisterRiemann I use another terminology, that is 1) convergent: $a_n to Lin mathbb{R}$, 2) divergent: $a_n toinfty$ or $a_n to-infty$ 3) Not convergent nor divergent otherwise.
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:15






  • 1




    $begingroup$
    That is a nonstandard terminology then. [1] [2]
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:31








  • 1




    $begingroup$
    @gimusi I've never seen such a convention. In any case, even the attached link classifies any sequence that does not converge is as a divergent sequence.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:42
















0












0








0





$begingroup$


"A bounded sequence cannot be divergent." Is this statement true?
As far as I know a bounded sequence can either be convergent or finitely oscillating, it cannot be divergent since it cannot diverge to infinity being a bounded sequence. Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?






real-analysis calculus sequences-and-series







share|cite|improve this question









$endgroup$




"A bounded sequence cannot be divergent." Is this statement true?
As far as I know a bounded sequence can either be convergent or finitely oscillating, it cannot be divergent since it cannot diverge to infinity being a bounded sequence. Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?






real-analysis calculus sequences-and-series




real-analysis calculus sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 14:08









jirenjiren

766




766








  • 3




    $begingroup$
    "Divergent" means "not convergent".
    $endgroup$
    – Hans Engler
    Dec 4 '18 at 14:09






  • 3




    $begingroup$
    Recall that a sequence is said to be divergent if it is not convergent. Hence if it oscillates, it is considered to be divergent.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:10










  • $begingroup$
    @MisterRiemann I use another terminology, that is 1) convergent: $a_n to Lin mathbb{R}$, 2) divergent: $a_n toinfty$ or $a_n to-infty$ 3) Not convergent nor divergent otherwise.
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:15






  • 1




    $begingroup$
    That is a nonstandard terminology then. [1] [2]
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:31








  • 1




    $begingroup$
    @gimusi I've never seen such a convention. In any case, even the attached link classifies any sequence that does not converge is as a divergent sequence.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:42
















  • 3




    $begingroup$
    "Divergent" means "not convergent".
    $endgroup$
    – Hans Engler
    Dec 4 '18 at 14:09






  • 3




    $begingroup$
    Recall that a sequence is said to be divergent if it is not convergent. Hence if it oscillates, it is considered to be divergent.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:10










  • $begingroup$
    @MisterRiemann I use another terminology, that is 1) convergent: $a_n to Lin mathbb{R}$, 2) divergent: $a_n toinfty$ or $a_n to-infty$ 3) Not convergent nor divergent otherwise.
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:15






  • 1




    $begingroup$
    That is a nonstandard terminology then. [1] [2]
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:31








  • 1




    $begingroup$
    @gimusi I've never seen such a convention. In any case, even the attached link classifies any sequence that does not converge is as a divergent sequence.
    $endgroup$
    – MisterRiemann
    Dec 4 '18 at 14:42










3




3




$begingroup$
"Divergent" means "not convergent".
$endgroup$
– Hans Engler
Dec 4 '18 at 14:09




$begingroup$
"Divergent" means "not convergent".
$endgroup$
– Hans Engler
Dec 4 '18 at 14:09




3




3




$begingroup$
Recall that a sequence is said to be divergent if it is not convergent. Hence if it oscillates, it is considered to be divergent.
$endgroup$
– MisterRiemann
Dec 4 '18 at 14:10




$begingroup$
Recall that a sequence is said to be divergent if it is not convergent. Hence if it oscillates, it is considered to be divergent.
$endgroup$
– MisterRiemann
Dec 4 '18 at 14:10












$begingroup$
@MisterRiemann I use another terminology, that is 1) convergent: $a_n to Lin mathbb{R}$, 2) divergent: $a_n toinfty$ or $a_n to-infty$ 3) Not convergent nor divergent otherwise.
$endgroup$
– gimusi
Dec 4 '18 at 14:15




$begingroup$
@MisterRiemann I use another terminology, that is 1) convergent: $a_n to Lin mathbb{R}$, 2) divergent: $a_n toinfty$ or $a_n to-infty$ 3) Not convergent nor divergent otherwise.
$endgroup$
– gimusi
Dec 4 '18 at 14:15




1




1




$begingroup$
That is a nonstandard terminology then. [1] [2]
$endgroup$
– MisterRiemann
Dec 4 '18 at 14:31






$begingroup$
That is a nonstandard terminology then. [1] [2]
$endgroup$
– MisterRiemann
Dec 4 '18 at 14:31






1




1




$begingroup$
@gimusi I've never seen such a convention. In any case, even the attached link classifies any sequence that does not converge is as a divergent sequence.
$endgroup$
– MisterRiemann
Dec 4 '18 at 14:42






$begingroup$
@gimusi I've never seen such a convention. In any case, even the attached link classifies any sequence that does not converge is as a divergent sequence.
$endgroup$
– MisterRiemann
Dec 4 '18 at 14:42












2 Answers
2






active

oldest

votes


















5












$begingroup$

$$ (-1)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What should that prove?
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:13










  • $begingroup$
    this sequence is bounded but divergent
    $endgroup$
    – staedtlerr
    Dec 4 '18 at 14:36










  • $begingroup$
    Yes sorry I was referring to a difefrent terminology! Of course your example is fine
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:45










  • $begingroup$
    I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
    $endgroup$
    – Michael
    Dec 4 '18 at 15:43



















-1












$begingroup$

Yes you can.



State that your sequence is unbounded. Suppose it has no upper bound (the same logic is to be applied if it has no lower bound).



So:



$$forall x in mathbb R, exists n in mathbb N, u_n > x$$



Work a bit with this, and the definition of limit:



$$ exists a in mathbb R, forall epsilon in mathbb R, exists n_0 in mathbb N, n > n_0 implies |u_n - a | < epsilon $$



The definition of limits tells you that for every $epsilon$, there is a rank $n_0$ such that $(u_n)_n$ is bounded for $n>n_0$. Since there is a finite number of terms $u_n$ for $0 le nle n_0$, the sequece is also bounded for $n le n_0$.



This is a contradiction with the undounded hypothesis, and you can then conclude that the "convergent" hypothesys is wrong.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think you've mixed up the pieces - this is a proof that a convergent sequence is bounded.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 14:44










  • $begingroup$
    @T.Bongers So, a non bounded sequence is non convergent. Which I though was the question, but it may have been a bit more precise than that, with "divergent or infinitely oscillating" not being equivalent to "non convergent", as shown by Michael Hoppe comment
    $endgroup$
    – F.Carette
    Dec 4 '18 at 14:54










  • $begingroup$
    Just from the title, though, the question is asking for an example of a bounded sequence which is divergent or a proof that a bounded sequence is convergent.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 15:01










  • $begingroup$
    @T.Bongers my bad, I was answering the last sentence: "Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?". I think I mixed up quite a few things indeed. I'll let the answer anyway, unless OP states that it doesn't help him at all. Thanks for pointing it out
    $endgroup$
    – F.Carette
    Dec 4 '18 at 15:11











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025626%2fa-bounded-sequence-cannot-be-divergent-true-or-false%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

$$ (-1)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What should that prove?
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:13










  • $begingroup$
    this sequence is bounded but divergent
    $endgroup$
    – staedtlerr
    Dec 4 '18 at 14:36










  • $begingroup$
    Yes sorry I was referring to a difefrent terminology! Of course your example is fine
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:45










  • $begingroup$
    I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
    $endgroup$
    – Michael
    Dec 4 '18 at 15:43
















5












$begingroup$

$$ (-1)^n$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What should that prove?
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:13










  • $begingroup$
    this sequence is bounded but divergent
    $endgroup$
    – staedtlerr
    Dec 4 '18 at 14:36










  • $begingroup$
    Yes sorry I was referring to a difefrent terminology! Of course your example is fine
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:45










  • $begingroup$
    I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
    $endgroup$
    – Michael
    Dec 4 '18 at 15:43














5












5








5





$begingroup$

$$ (-1)^n$$






share|cite|improve this answer









$endgroup$



$$ (-1)^n$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 14:12









MathematicsStudent1122MathematicsStudent1122

8,67622467




8,67622467












  • $begingroup$
    What should that prove?
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:13










  • $begingroup$
    this sequence is bounded but divergent
    $endgroup$
    – staedtlerr
    Dec 4 '18 at 14:36










  • $begingroup$
    Yes sorry I was referring to a difefrent terminology! Of course your example is fine
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:45










  • $begingroup$
    I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
    $endgroup$
    – Michael
    Dec 4 '18 at 15:43


















  • $begingroup$
    What should that prove?
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:13










  • $begingroup$
    this sequence is bounded but divergent
    $endgroup$
    – staedtlerr
    Dec 4 '18 at 14:36










  • $begingroup$
    Yes sorry I was referring to a difefrent terminology! Of course your example is fine
    $endgroup$
    – gimusi
    Dec 4 '18 at 14:45










  • $begingroup$
    I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
    $endgroup$
    – Michael
    Dec 4 '18 at 15:43
















$begingroup$
What should that prove?
$endgroup$
– gimusi
Dec 4 '18 at 14:13




$begingroup$
What should that prove?
$endgroup$
– gimusi
Dec 4 '18 at 14:13












$begingroup$
this sequence is bounded but divergent
$endgroup$
– staedtlerr
Dec 4 '18 at 14:36




$begingroup$
this sequence is bounded but divergent
$endgroup$
– staedtlerr
Dec 4 '18 at 14:36












$begingroup$
Yes sorry I was referring to a difefrent terminology! Of course your example is fine
$endgroup$
– gimusi
Dec 4 '18 at 14:45




$begingroup$
Yes sorry I was referring to a difefrent terminology! Of course your example is fine
$endgroup$
– gimusi
Dec 4 '18 at 14:45












$begingroup$
I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
$endgroup$
– Michael
Dec 4 '18 at 15:43




$begingroup$
I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
$endgroup$
– Michael
Dec 4 '18 at 15:43











-1












$begingroup$

Yes you can.



State that your sequence is unbounded. Suppose it has no upper bound (the same logic is to be applied if it has no lower bound).



So:



$$forall x in mathbb R, exists n in mathbb N, u_n > x$$



Work a bit with this, and the definition of limit:



$$ exists a in mathbb R, forall epsilon in mathbb R, exists n_0 in mathbb N, n > n_0 implies |u_n - a | < epsilon $$



The definition of limits tells you that for every $epsilon$, there is a rank $n_0$ such that $(u_n)_n$ is bounded for $n>n_0$. Since there is a finite number of terms $u_n$ for $0 le nle n_0$, the sequece is also bounded for $n le n_0$.



This is a contradiction with the undounded hypothesis, and you can then conclude that the "convergent" hypothesys is wrong.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think you've mixed up the pieces - this is a proof that a convergent sequence is bounded.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 14:44










  • $begingroup$
    @T.Bongers So, a non bounded sequence is non convergent. Which I though was the question, but it may have been a bit more precise than that, with "divergent or infinitely oscillating" not being equivalent to "non convergent", as shown by Michael Hoppe comment
    $endgroup$
    – F.Carette
    Dec 4 '18 at 14:54










  • $begingroup$
    Just from the title, though, the question is asking for an example of a bounded sequence which is divergent or a proof that a bounded sequence is convergent.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 15:01










  • $begingroup$
    @T.Bongers my bad, I was answering the last sentence: "Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?". I think I mixed up quite a few things indeed. I'll let the answer anyway, unless OP states that it doesn't help him at all. Thanks for pointing it out
    $endgroup$
    – F.Carette
    Dec 4 '18 at 15:11
















-1












$begingroup$

Yes you can.



State that your sequence is unbounded. Suppose it has no upper bound (the same logic is to be applied if it has no lower bound).



So:



$$forall x in mathbb R, exists n in mathbb N, u_n > x$$



Work a bit with this, and the definition of limit:



$$ exists a in mathbb R, forall epsilon in mathbb R, exists n_0 in mathbb N, n > n_0 implies |u_n - a | < epsilon $$



The definition of limits tells you that for every $epsilon$, there is a rank $n_0$ such that $(u_n)_n$ is bounded for $n>n_0$. Since there is a finite number of terms $u_n$ for $0 le nle n_0$, the sequece is also bounded for $n le n_0$.



This is a contradiction with the undounded hypothesis, and you can then conclude that the "convergent" hypothesys is wrong.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think you've mixed up the pieces - this is a proof that a convergent sequence is bounded.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 14:44










  • $begingroup$
    @T.Bongers So, a non bounded sequence is non convergent. Which I though was the question, but it may have been a bit more precise than that, with "divergent or infinitely oscillating" not being equivalent to "non convergent", as shown by Michael Hoppe comment
    $endgroup$
    – F.Carette
    Dec 4 '18 at 14:54










  • $begingroup$
    Just from the title, though, the question is asking for an example of a bounded sequence which is divergent or a proof that a bounded sequence is convergent.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 15:01










  • $begingroup$
    @T.Bongers my bad, I was answering the last sentence: "Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?". I think I mixed up quite a few things indeed. I'll let the answer anyway, unless OP states that it doesn't help him at all. Thanks for pointing it out
    $endgroup$
    – F.Carette
    Dec 4 '18 at 15:11














-1












-1








-1





$begingroup$

Yes you can.



State that your sequence is unbounded. Suppose it has no upper bound (the same logic is to be applied if it has no lower bound).



So:



$$forall x in mathbb R, exists n in mathbb N, u_n > x$$



Work a bit with this, and the definition of limit:



$$ exists a in mathbb R, forall epsilon in mathbb R, exists n_0 in mathbb N, n > n_0 implies |u_n - a | < epsilon $$



The definition of limits tells you that for every $epsilon$, there is a rank $n_0$ such that $(u_n)_n$ is bounded for $n>n_0$. Since there is a finite number of terms $u_n$ for $0 le nle n_0$, the sequece is also bounded for $n le n_0$.



This is a contradiction with the undounded hypothesis, and you can then conclude that the "convergent" hypothesys is wrong.






share|cite|improve this answer









$endgroup$



Yes you can.



State that your sequence is unbounded. Suppose it has no upper bound (the same logic is to be applied if it has no lower bound).



So:



$$forall x in mathbb R, exists n in mathbb N, u_n > x$$



Work a bit with this, and the definition of limit:



$$ exists a in mathbb R, forall epsilon in mathbb R, exists n_0 in mathbb N, n > n_0 implies |u_n - a | < epsilon $$



The definition of limits tells you that for every $epsilon$, there is a rank $n_0$ such that $(u_n)_n$ is bounded for $n>n_0$. Since there is a finite number of terms $u_n$ for $0 le nle n_0$, the sequece is also bounded for $n le n_0$.



This is a contradiction with the undounded hypothesis, and you can then conclude that the "convergent" hypothesys is wrong.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 14:26









F.CaretteF.Carette

1,21612




1,21612








  • 1




    $begingroup$
    I think you've mixed up the pieces - this is a proof that a convergent sequence is bounded.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 14:44










  • $begingroup$
    @T.Bongers So, a non bounded sequence is non convergent. Which I though was the question, but it may have been a bit more precise than that, with "divergent or infinitely oscillating" not being equivalent to "non convergent", as shown by Michael Hoppe comment
    $endgroup$
    – F.Carette
    Dec 4 '18 at 14:54










  • $begingroup$
    Just from the title, though, the question is asking for an example of a bounded sequence which is divergent or a proof that a bounded sequence is convergent.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 15:01










  • $begingroup$
    @T.Bongers my bad, I was answering the last sentence: "Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?". I think I mixed up quite a few things indeed. I'll let the answer anyway, unless OP states that it doesn't help him at all. Thanks for pointing it out
    $endgroup$
    – F.Carette
    Dec 4 '18 at 15:11














  • 1




    $begingroup$
    I think you've mixed up the pieces - this is a proof that a convergent sequence is bounded.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 14:44










  • $begingroup$
    @T.Bongers So, a non bounded sequence is non convergent. Which I though was the question, but it may have been a bit more precise than that, with "divergent or infinitely oscillating" not being equivalent to "non convergent", as shown by Michael Hoppe comment
    $endgroup$
    – F.Carette
    Dec 4 '18 at 14:54










  • $begingroup$
    Just from the title, though, the question is asking for an example of a bounded sequence which is divergent or a proof that a bounded sequence is convergent.
    $endgroup$
    – T. Bongers
    Dec 4 '18 at 15:01










  • $begingroup$
    @T.Bongers my bad, I was answering the last sentence: "Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?". I think I mixed up quite a few things indeed. I'll let the answer anyway, unless OP states that it doesn't help him at all. Thanks for pointing it out
    $endgroup$
    – F.Carette
    Dec 4 '18 at 15:11








1




1




$begingroup$
I think you've mixed up the pieces - this is a proof that a convergent sequence is bounded.
$endgroup$
– T. Bongers
Dec 4 '18 at 14:44




$begingroup$
I think you've mixed up the pieces - this is a proof that a convergent sequence is bounded.
$endgroup$
– T. Bongers
Dec 4 '18 at 14:44












$begingroup$
@T.Bongers So, a non bounded sequence is non convergent. Which I though was the question, but it may have been a bit more precise than that, with "divergent or infinitely oscillating" not being equivalent to "non convergent", as shown by Michael Hoppe comment
$endgroup$
– F.Carette
Dec 4 '18 at 14:54




$begingroup$
@T.Bongers So, a non bounded sequence is non convergent. Which I though was the question, but it may have been a bit more precise than that, with "divergent or infinitely oscillating" not being equivalent to "non convergent", as shown by Michael Hoppe comment
$endgroup$
– F.Carette
Dec 4 '18 at 14:54












$begingroup$
Just from the title, though, the question is asking for an example of a bounded sequence which is divergent or a proof that a bounded sequence is convergent.
$endgroup$
– T. Bongers
Dec 4 '18 at 15:01




$begingroup$
Just from the title, though, the question is asking for an example of a bounded sequence which is divergent or a proof that a bounded sequence is convergent.
$endgroup$
– T. Bongers
Dec 4 '18 at 15:01












$begingroup$
@T.Bongers my bad, I was answering the last sentence: "Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?". I think I mixed up quite a few things indeed. I'll let the answer anyway, unless OP states that it doesn't help him at all. Thanks for pointing it out
$endgroup$
– F.Carette
Dec 4 '18 at 15:11




$begingroup$
@T.Bongers my bad, I was answering the last sentence: "Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?". I think I mixed up quite a few things indeed. I'll let the answer anyway, unless OP states that it doesn't help him at all. Thanks for pointing it out
$endgroup$
– F.Carette
Dec 4 '18 at 15:11


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025626%2fa-bounded-sequence-cannot-be-divergent-true-or-false%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more