Can a bounded number sequence be strictly ascending?Bounded Sequence ProofEvery bounded monotone sequence convergesCauchy sequences are bounded?!Why is sequence $(1+frac1n)^n+1$ descending?Show that for a sequence $p_n$ of real numbers, $limsup p_n < +∞$ iff $p_n$ is bounded above.If $a_n$ and $b_n$ are equivalent sequences and $a_n$ is bounded then so is $b_n$.Show that $s_n$ is bounded if $s_n = b_1r + b_2r^2 + … + b_nr^n$ and $0 < r < 1$.Implicit Fact about Bounded Monotone Sequence convergesBolzano-Weierstrass boundedProve that if a sequence is eventually bounded then it is bounded
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Can a bounded number sequence be strictly ascending?
Bounded Sequence ProofEvery bounded monotone sequence convergesCauchy sequences are bounded?!Why is sequence $(1+frac1n)^n+1$ descending?Show that for a sequence $p_n$ of real numbers, $limsup p_n < +∞$ iff $p_n$ is bounded above.If $a_n$ and $b_n$ are equivalent sequences and $a_n$ is bounded then so is $b_n$.Show that $s_n$ is bounded if $s_n = b_1r + b_2r^2 + … + b_nr^n$ and $0 < r < 1$.Implicit Fact about Bounded Monotone Sequence convergesBolzano-Weierstrass boundedProve that if a sequence is eventually bounded then it is bounded
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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 10 hours ago
furfurfurfur
719
$endgroup$
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 10 hours ago
furfurfurfur
719
$endgroup$
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
10 hours ago
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
7 hours ago
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 10 hours ago
furfurfurfur
719
$endgroup$
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 10 hours ago
furfurfurfur
719
asked 10 hours ago
furfurfurfur
719
edited 10 hours ago
furfur
asked 10 hours ago
furfurfurfur
719
asked 10 hours ago
furfurfurfur
719
asked 10 hours ago
furfurfurfur
719
719
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
10 hours ago
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
7 hours ago
add a comment |
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
10 hours ago
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
7 hours ago
1
1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
10 hours ago
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
10 hours ago
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
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered 10 hours ago
StackTDStackTD
23.5k2154
$endgroup$
add a comment |
$begingroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered 10 hours ago
Robert IsraelRobert Israel
327k23216470
$endgroup$
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 the base for Zeno's Paradoxes
answered 6 hours ago
ntgntg
1436
$endgroup$
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
5 hours ago
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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active
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$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered 10 hours ago
StackTDStackTD
23.5k2154
$endgroup$
add a comment |
$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered 10 hours ago
StackTDStackTD
23.5k2154
$endgroup$
add a comment |
$begingroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered 10 hours ago
StackTDStackTD
23.5k2154
$endgroup$
Can a bounded number sequence be strictly ascending?
Sure it can.
Hint
$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$
answered 10 hours ago
StackTDStackTD
23.5k2154
edited 4 hours ago
answered 10 hours ago
StackTDStackTD
23.5k2154
answered 10 hours ago
StackTDStackTD
23.5k2154
answered 10 hours ago
StackTDStackTD
23.5k2154
23.5k2154
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 10 hours ago
Robert IsraelRobert Israel
327k23216470
$endgroup$
add a comment |
$begingroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered 10 hours ago
Robert IsraelRobert Israel
327k23216470
$endgroup$
add a comment |
$begingroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered 10 hours ago
Robert IsraelRobert Israel
327k23216470
$endgroup$
I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.
answered 10 hours ago
Robert IsraelRobert Israel
327k23216470
answered 10 hours ago
Robert IsraelRobert Israel
327k23216470
answered 10 hours ago
Robert IsraelRobert Israel
327k23216470
answered 10 hours ago
Robert IsraelRobert Israel
327k23216470
327k23216470
add a comment |
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 the base for Zeno's Paradoxes
answered 6 hours ago
ntgntg
1436
$endgroup$
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
5 hours ago
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 the base for Zeno's Paradoxes
answered 6 hours ago
ntgntg
1436
$endgroup$
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
5 hours ago
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 the base for Zeno's Paradoxes
answered 6 hours ago
ntgntg
1436
$endgroup$
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 the base for Zeno's Paradoxes
answered 6 hours ago
ntgntg
1436
answered 6 hours ago
ntgntg
1436
answered 6 hours ago
ntgntg
1436
answered 6 hours ago
ntgntg
1436
1436
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
5 hours ago
add a comment |
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
5 hours ago
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
5 hours ago
$begingroup$
+1 for Zeno's paradox
$endgroup$
– Pere
5 hours ago
add a comment |
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1
$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
10 hours ago
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
7 hours ago