Can one construct a convergent series out of a non-decreasing Cauchy sequence in $mathbb{R}$ [closed]
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If we have a non-decreasing Cauchy sequence of non-negative real numbers ${x_n}$, will the sequence of partial sums ${S_n}_{n=1}^{infty}$ where $S_n= x_1 + x_2 +..+x_n$ converges?
real-analysis sequences-and-series convergence
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If we have a non-decreasing Cauchy sequence of non-negative real numbers ${x_n}$, will the sequence of partial sums ${S_n}_{n=1}^{infty}$ where $S_n= x_1 + x_2 +..+x_n$ converges?
real-analysis sequences-and-series convergence
closed as off-topic by Did, amWhy, Shailesh, max_zorn, Rebellos Nov 18 at 11:59
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, amWhy, Shailesh, max_zorn, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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If we have a non-decreasing Cauchy sequence of non-negative real numbers ${x_n}$, will the sequence of partial sums ${S_n}_{n=1}^{infty}$ where $S_n= x_1 + x_2 +..+x_n$ converges?
real-analysis sequences-and-series convergence
If we have a non-decreasing Cauchy sequence of non-negative real numbers ${x_n}$, will the sequence of partial sums ${S_n}_{n=1}^{infty}$ where $S_n= x_1 + x_2 +..+x_n$ converges?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Nov 17 at 15:28
asked Nov 17 at 13:00
HybridAlien
2008
2008
closed as off-topic by Did, amWhy, Shailesh, max_zorn, Rebellos Nov 18 at 11:59
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, amWhy, Shailesh, max_zorn, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, amWhy, Shailesh, max_zorn, Rebellos Nov 18 at 11:59
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, amWhy, Shailesh, max_zorn, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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3 Answers
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Suppose $S_n$ converges, it implies $x_nto 0$
Since we have $x_nge 0$ then the sequence will be globally decreasing (e.g $forall n exists n_0>n mid forall i> n_0,,x_ile x_{n_0}$)
But the sequence can be non-decreasing while being globally decreasing. It suffice for instance to interleave two decreasing sequences.
$x_{2n} = dfrac{1}{n^alpha}$ and $x_{2n+1}=dfrac{2}{n^alpha}$ for instance.
We have $(x_{2n})_nsearrow$ and $(x_{2n+1})_nsearrow$ but $x_{2n+1}>x_{2n}$ so $(x_n)_n$ is non-decreasing.
Depending on the value of $alpha$ the partial sum $S_n$ will converge or diverge.
So the answer for the TITLE is YES, you can construct such series, take $alpha=2$ for instance.
And the answer for the wording below is NOT NECESSARILY, take $alpha=1$ for instance.
Nice answer Zwim!!
– BAYMAX
Nov 17 at 16:19
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No. All that being a Cauchy sequence tells you is that the terms eventually get very close to each other. However, the could do this in the vicinity of $5$, for example $a_n = 5 + dfrac 1n$. Then the sum is $ge 5+5+5 + cdots$, which diverges.
Well $a_nsearrow$, and OP asks explicitly for non-decreasing sequences...
– zwim
Nov 17 at 15:55
1
@zwim Oh sorry forgot. Anyways it's wasy to modify this example to $5-1/n$ whose sum still diverges.
– Ovi
Nov 17 at 15:58
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No. All convergent sequences are Cauchy, but that doesn't mean that a series of positive terms which tend to $0$ (hence the sequence of these terms is Cauchy) is itself Cauchy, as this would imply that a series with positive terms converges if and only if its general tends to $0$, which false in $mathbf R$ (but is true in the field of $p$-adic numbers).
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Suppose $S_n$ converges, it implies $x_nto 0$
Since we have $x_nge 0$ then the sequence will be globally decreasing (e.g $forall n exists n_0>n mid forall i> n_0,,x_ile x_{n_0}$)
But the sequence can be non-decreasing while being globally decreasing. It suffice for instance to interleave two decreasing sequences.
$x_{2n} = dfrac{1}{n^alpha}$ and $x_{2n+1}=dfrac{2}{n^alpha}$ for instance.
We have $(x_{2n})_nsearrow$ and $(x_{2n+1})_nsearrow$ but $x_{2n+1}>x_{2n}$ so $(x_n)_n$ is non-decreasing.
Depending on the value of $alpha$ the partial sum $S_n$ will converge or diverge.
So the answer for the TITLE is YES, you can construct such series, take $alpha=2$ for instance.
And the answer for the wording below is NOT NECESSARILY, take $alpha=1$ for instance.
Nice answer Zwim!!
– BAYMAX
Nov 17 at 16:19
add a comment |
up vote
2
down vote
accepted
Suppose $S_n$ converges, it implies $x_nto 0$
Since we have $x_nge 0$ then the sequence will be globally decreasing (e.g $forall n exists n_0>n mid forall i> n_0,,x_ile x_{n_0}$)
But the sequence can be non-decreasing while being globally decreasing. It suffice for instance to interleave two decreasing sequences.
$x_{2n} = dfrac{1}{n^alpha}$ and $x_{2n+1}=dfrac{2}{n^alpha}$ for instance.
We have $(x_{2n})_nsearrow$ and $(x_{2n+1})_nsearrow$ but $x_{2n+1}>x_{2n}$ so $(x_n)_n$ is non-decreasing.
Depending on the value of $alpha$ the partial sum $S_n$ will converge or diverge.
So the answer for the TITLE is YES, you can construct such series, take $alpha=2$ for instance.
And the answer for the wording below is NOT NECESSARILY, take $alpha=1$ for instance.
Nice answer Zwim!!
– BAYMAX
Nov 17 at 16:19
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose $S_n$ converges, it implies $x_nto 0$
Since we have $x_nge 0$ then the sequence will be globally decreasing (e.g $forall n exists n_0>n mid forall i> n_0,,x_ile x_{n_0}$)
But the sequence can be non-decreasing while being globally decreasing. It suffice for instance to interleave two decreasing sequences.
$x_{2n} = dfrac{1}{n^alpha}$ and $x_{2n+1}=dfrac{2}{n^alpha}$ for instance.
We have $(x_{2n})_nsearrow$ and $(x_{2n+1})_nsearrow$ but $x_{2n+1}>x_{2n}$ so $(x_n)_n$ is non-decreasing.
Depending on the value of $alpha$ the partial sum $S_n$ will converge or diverge.
So the answer for the TITLE is YES, you can construct such series, take $alpha=2$ for instance.
And the answer for the wording below is NOT NECESSARILY, take $alpha=1$ for instance.
Suppose $S_n$ converges, it implies $x_nto 0$
Since we have $x_nge 0$ then the sequence will be globally decreasing (e.g $forall n exists n_0>n mid forall i> n_0,,x_ile x_{n_0}$)
But the sequence can be non-decreasing while being globally decreasing. It suffice for instance to interleave two decreasing sequences.
$x_{2n} = dfrac{1}{n^alpha}$ and $x_{2n+1}=dfrac{2}{n^alpha}$ for instance.
We have $(x_{2n})_nsearrow$ and $(x_{2n+1})_nsearrow$ but $x_{2n+1}>x_{2n}$ so $(x_n)_n$ is non-decreasing.
Depending on the value of $alpha$ the partial sum $S_n$ will converge or diverge.
So the answer for the TITLE is YES, you can construct such series, take $alpha=2$ for instance.
And the answer for the wording below is NOT NECESSARILY, take $alpha=1$ for instance.
answered Nov 17 at 15:53
zwim
11.2k628
11.2k628
Nice answer Zwim!!
– BAYMAX
Nov 17 at 16:19
add a comment |
Nice answer Zwim!!
– BAYMAX
Nov 17 at 16:19
Nice answer Zwim!!
– BAYMAX
Nov 17 at 16:19
Nice answer Zwim!!
– BAYMAX
Nov 17 at 16:19
add a comment |
up vote
1
down vote
No. All that being a Cauchy sequence tells you is that the terms eventually get very close to each other. However, the could do this in the vicinity of $5$, for example $a_n = 5 + dfrac 1n$. Then the sum is $ge 5+5+5 + cdots$, which diverges.
Well $a_nsearrow$, and OP asks explicitly for non-decreasing sequences...
– zwim
Nov 17 at 15:55
1
@zwim Oh sorry forgot. Anyways it's wasy to modify this example to $5-1/n$ whose sum still diverges.
– Ovi
Nov 17 at 15:58
add a comment |
up vote
1
down vote
No. All that being a Cauchy sequence tells you is that the terms eventually get very close to each other. However, the could do this in the vicinity of $5$, for example $a_n = 5 + dfrac 1n$. Then the sum is $ge 5+5+5 + cdots$, which diverges.
Well $a_nsearrow$, and OP asks explicitly for non-decreasing sequences...
– zwim
Nov 17 at 15:55
1
@zwim Oh sorry forgot. Anyways it's wasy to modify this example to $5-1/n$ whose sum still diverges.
– Ovi
Nov 17 at 15:58
add a comment |
up vote
1
down vote
up vote
1
down vote
No. All that being a Cauchy sequence tells you is that the terms eventually get very close to each other. However, the could do this in the vicinity of $5$, for example $a_n = 5 + dfrac 1n$. Then the sum is $ge 5+5+5 + cdots$, which diverges.
No. All that being a Cauchy sequence tells you is that the terms eventually get very close to each other. However, the could do this in the vicinity of $5$, for example $a_n = 5 + dfrac 1n$. Then the sum is $ge 5+5+5 + cdots$, which diverges.
answered Nov 17 at 15:32
Ovi
12.1k938108
12.1k938108
Well $a_nsearrow$, and OP asks explicitly for non-decreasing sequences...
– zwim
Nov 17 at 15:55
1
@zwim Oh sorry forgot. Anyways it's wasy to modify this example to $5-1/n$ whose sum still diverges.
– Ovi
Nov 17 at 15:58
add a comment |
Well $a_nsearrow$, and OP asks explicitly for non-decreasing sequences...
– zwim
Nov 17 at 15:55
1
@zwim Oh sorry forgot. Anyways it's wasy to modify this example to $5-1/n$ whose sum still diverges.
– Ovi
Nov 17 at 15:58
Well $a_nsearrow$, and OP asks explicitly for non-decreasing sequences...
– zwim
Nov 17 at 15:55
Well $a_nsearrow$, and OP asks explicitly for non-decreasing sequences...
– zwim
Nov 17 at 15:55
1
1
@zwim Oh sorry forgot. Anyways it's wasy to modify this example to $5-1/n$ whose sum still diverges.
– Ovi
Nov 17 at 15:58
@zwim Oh sorry forgot. Anyways it's wasy to modify this example to $5-1/n$ whose sum still diverges.
– Ovi
Nov 17 at 15:58
add a comment |
up vote
0
down vote
No. All convergent sequences are Cauchy, but that doesn't mean that a series of positive terms which tend to $0$ (hence the sequence of these terms is Cauchy) is itself Cauchy, as this would imply that a series with positive terms converges if and only if its general tends to $0$, which false in $mathbf R$ (but is true in the field of $p$-adic numbers).
add a comment |
up vote
0
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No. All convergent sequences are Cauchy, but that doesn't mean that a series of positive terms which tend to $0$ (hence the sequence of these terms is Cauchy) is itself Cauchy, as this would imply that a series with positive terms converges if and only if its general tends to $0$, which false in $mathbf R$ (but is true in the field of $p$-adic numbers).
add a comment |
up vote
0
down vote
up vote
0
down vote
No. All convergent sequences are Cauchy, but that doesn't mean that a series of positive terms which tend to $0$ (hence the sequence of these terms is Cauchy) is itself Cauchy, as this would imply that a series with positive terms converges if and only if its general tends to $0$, which false in $mathbf R$ (but is true in the field of $p$-adic numbers).
No. All convergent sequences are Cauchy, but that doesn't mean that a series of positive terms which tend to $0$ (hence the sequence of these terms is Cauchy) is itself Cauchy, as this would imply that a series with positive terms converges if and only if its general tends to $0$, which false in $mathbf R$ (but is true in the field of $p$-adic numbers).
answered Nov 17 at 13:16
Bernard
116k637108
116k637108
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