Can a bounded number sequence be strictly ascending? [closed]
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The title says it. Can a bounded number sequence be strictly ascending / descending?
I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.
sequences-and-series
asked Mar 12 at 12:14
furfurfurfur
919
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closed as off-topic by user21820, Song, Alex Provost, RRL, Parcly Taxel Mar 16 at 15:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Song, Alex Provost, RRL, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The title says it. Can a bounded number sequence be strictly ascending / descending?
I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.
sequences-and-series
asked Mar 12 at 12:14
furfurfurfur
919
$endgroup$
closed as off-topic by user21820, Song, Alex Provost, RRL, Parcly Taxel Mar 16 at 15:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Song, Alex Provost, RRL, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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A series deals with summation. A sequence deals with individual elements.
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– Subhasis Biswas
Mar 12 at 12:19
4
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You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
Mar 12 at 15:12
add a comment |
$begingroup$
The title says it. Can a bounded number sequence be strictly ascending / descending?
I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.
sequences-and-series
asked Mar 12 at 12:14
furfurfurfur
919
$endgroup$
The title says it. Can a bounded number sequence be strictly ascending / descending?
I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.
sequences-and-series
sequences-and-series
asked Mar 12 at 12:14
furfurfurfur
919
asked Mar 12 at 12:14
furfurfurfur
919
edited Mar 12 at 12:26
furfur
asked Mar 12 at 12:14
furfurfurfur
919
asked Mar 12 at 12:14
furfurfurfur
919
asked Mar 12 at 12:14
furfurfurfur
919
919
closed as off-topic by user21820, Song, Alex Provost, RRL, Parcly Taxel Mar 16 at 15:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Song, Alex Provost, RRL, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Song, Alex Provost, RRL, Parcly Taxel Mar 16 at 15:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Song, Alex Provost, RRL, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
Mar 12 at 12:19
4
$begingroup$
You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
Mar 12 at 15:12
add a comment |
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
Mar 12 at 12:19
4
$begingroup$
You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
Mar 12 at 15:12
1
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
Mar 12 at 12:19
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
Mar 12 at 12:19
4
4
$begingroup$
You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
Mar 12 at 15:12
$begingroup$
You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
Mar 12 at 15:12
add a comment |
3 Answers
3
active
oldest
votes
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Yes.
Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"
So basically the sequence of the partial sums of e.g. a geometric series
with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:
This is also relates to Zeno's Paradoxes.
answered Mar 12 at 16:21
ntgntg
1887
$endgroup$
$begingroup$
+1 for Zeno's paradox
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– Pere
Mar 12 at 16:57
add a comment |
$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes.
Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"
So basically the sequence of the partial sums of e.g. a geometric series
with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:
This is also relates to Zeno's Paradoxes.
answered Mar 12 at 16:21
ntgntg
1887
$endgroup$
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
Mar 12 at 16:57
add a comment |
$begingroup$
Yes.
Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"
So basically the sequence of the partial sums of e.g. a geometric series
with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:
This is also relates to Zeno's Paradoxes.
answered Mar 12 at 16:21
ntgntg
1887
$endgroup$
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
Mar 12 at 16:57
add a comment |
$begingroup$
Yes.
Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"
So basically the sequence of the partial sums of e.g. a geometric series
with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:
This is also relates to Zeno's Paradoxes.
answered Mar 12 at 16:21
ntgntg
1887
$endgroup$
Yes.
Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"
So basically the sequence of the partial sums of e.g. a geometric series
with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:
This is also relates to Zeno's Paradoxes.
answered Mar 12 at 16:21
ntgntg
1887
edited Mar 14 at 9:19
answered Mar 12 at 16:21
ntgntg
1887
answered Mar 12 at 16:21
ntgntg
1887
answered Mar 12 at 16:21
ntgntg
1887
1887
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
Mar 12 at 16:57
add a comment |
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
Mar 12 at 16:57
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
Mar 12 at 16:57
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
Mar 12 at 16:57
add a comment |
$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
$endgroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
edited Mar 12 at 17:34
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
answered Mar 12 at 12:16
StackTDStackTD
24.3k2254
24.3k2254
add a comment |
add a comment |
$begingroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
$endgroup$
add a comment |
$begingroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
$endgroup$
add a comment |
$begingroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
$endgroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
answered Mar 12 at 12:16
Robert IsraelRobert Israel
332k23222479
332k23222479
add a comment |
add a comment |
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
Mar 12 at 12:19
4
$begingroup$
You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
Mar 12 at 15:12