A bounded sequence cannot be divergent. True or false
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"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
asked Dec 4 '18 at 14:08
jirenjiren
766
$endgroup$
|
show 6 more comments
$begingroup$
"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
asked Dec 4 '18 at 14:08
jirenjiren
766
$endgroup$
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.
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– MisterRiemann
Dec 4 '18 at 14:10
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@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.
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– gimusi
Dec 4 '18 at 14:15
1
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That is a nonstandard terminology then. [1] [2]
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– 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
|
show 6 more comments
$begingroup$
"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
asked Dec 4 '18 at 14:08
jirenjiren
766
$endgroup$
"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
asked Dec 4 '18 at 14:08
jirenjiren
766
asked Dec 4 '18 at 14:08
jirenjiren
766
asked Dec 4 '18 at 14:08
jirenjiren
766
asked Dec 4 '18 at 14:08
jirenjiren
766
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
|
show 6 more comments
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
|
show 6 more comments
2 Answers
2
active
oldest
votes
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$$ (-1)^n$$
answered Dec 4 '18 at 14:12
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$begingroup$
What should that prove?
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– gimusi
Dec 4 '18 at 14:13
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this sequence is bounded but divergent
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– staedtlerr
Dec 4 '18 at 14:36
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Yes sorry I was referring to a difefrent terminology! Of course your example is fine
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– gimusi
Dec 4 '18 at 14:45
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I am confused about the asker being "jiren" and also seemingly "gimusi" at the same time.
$endgroup$
– Michael
Dec 4 '18 at 15:43
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 14:26
F.CaretteF.Carette
1,21612
$endgroup$
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
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@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
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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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$ (-1)^n$$
answered Dec 4 '18 at 14:12
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$ (-1)^n$$
answered Dec 4 '18 at 14:12
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$ (-1)^n$$
answered Dec 4 '18 at 14:12
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$$ (-1)^n$$
answered Dec 4 '18 at 14:12
MathematicsStudent1122MathematicsStudent1122
8,67622467
answered Dec 4 '18 at 14:12
MathematicsStudent1122MathematicsStudent1122
8,67622467
answered Dec 4 '18 at 14:12
MathematicsStudent1122MathematicsStudent1122
8,67622467
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 14:26
F.CaretteF.Carette
1,21612
$endgroup$
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
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 14:26
F.CaretteF.Carette
1,21612
$endgroup$
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
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 14:26
F.CaretteF.Carette
1,21612
$endgroup$
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.
answered Dec 4 '18 at 14:26
F.CaretteF.Carette
1,21612
answered Dec 4 '18 at 14:26
F.CaretteF.Carette
1,21612
answered Dec 4 '18 at 14:26
F.CaretteF.Carette
1,21612
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
add a comment |
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
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– F.Carette
Dec 4 '18 at 15:11
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3
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"Divergent" means "not convergent".
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– Hans Engler
Dec 4 '18 at 14:09
3
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Recall that a sequence is said to be divergent if it is not convergent. Hence if it oscillates, it is considered to be divergent.
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– MisterRiemann
Dec 4 '18 at 14:10
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@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.
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– gimusi
Dec 4 '18 at 14:15
1
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That is a nonstandard terminology then. [1] [2]
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– MisterRiemann
Dec 4 '18 at 14:31
1
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@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.
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– MisterRiemann
Dec 4 '18 at 14:42