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







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$endgroup$








  • 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






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oldest

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$$ (-1)^n$$






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  • $begingroup$
    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











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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

8,62122467




8,62122467












  • $begingroup$
    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

1,22612




1,22612








  • 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


















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