Example of series such that every $sum_i=1^infty b_n_i$ converges but $sum_n=1^infty |b_n|$ diverges
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Does there exist a series $sum_n=1^infty b_n$ such that $sum_i=1^infty b_n_i$ converges for any $n_1 < n_2 < ldots$ but $sum_n=1^infty |b_n|$ diverges?
Certainly the example would have to be conditionally convergent, but such standard examples as the alternating harmonic series don't seem to work (at least from what I've managed to show).
Any ideas?
real-analysis sequences-and-series examples-counterexamples
asked Aug 26 at 23:36
CuriousKid7
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up vote
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Does there exist a series $sum_n=1^infty b_n$ such that $sum_i=1^infty b_n_i$ converges for any $n_1 < n_2 < ldots$ but $sum_n=1^infty |b_n|$ diverges?
Certainly the example would have to be conditionally convergent, but such standard examples as the alternating harmonic series don't seem to work (at least from what I've managed to show).
Any ideas?
real-analysis sequences-and-series examples-counterexamples
asked Aug 26 at 23:36
CuriousKid7
1,573617
If I don't misunderstand something, shouldn't be the index of the series over the $b_n_i$'s be $i$ and not $n$? Also, interesting question, +1.
– zzuussee
Aug 26 at 23:38
@zzuussee Fixed
– CuriousKid7
Aug 26 at 23:39
2
The sum consisting of all positive terms, $A = sum_i; b_igeq 0 b_i$ converges. Likewise for negative terms, $B = sum_j; b_j < 0 b_j$. Since $A-B = sum left | b_i right |$, $sum left | b_i right |$ must then converge.
– LPenguin
Aug 26 at 23:48
1
@SangchulLee But how do we know the rearrangement preserves the order of the $n_i$?
– CuriousKid7
Aug 26 at 23:49
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Does there exist a series $sum_n=1^infty b_n$ such that $sum_i=1^infty b_n_i$ converges for any $n_1 < n_2 < ldots$ but $sum_n=1^infty |b_n|$ diverges?
Certainly the example would have to be conditionally convergent, but such standard examples as the alternating harmonic series don't seem to work (at least from what I've managed to show).
Any ideas?
real-analysis sequences-and-series examples-counterexamples
asked Aug 26 at 23:36
CuriousKid7
1,573617
Does there exist a series $sum_n=1^infty b_n$ such that $sum_i=1^infty b_n_i$ converges for any $n_1 < n_2 < ldots$ but $sum_n=1^infty |b_n|$ diverges?
Certainly the example would have to be conditionally convergent, but such standard examples as the alternating harmonic series don't seem to work (at least from what I've managed to show).
Any ideas?
real-analysis sequences-and-series examples-counterexamples
asked Aug 26 at 23:36
CuriousKid7
1,573617
edited Aug 26 at 23:39
asked Aug 26 at 23:36
CuriousKid7
1,573617
asked Aug 26 at 23:36
CuriousKid7
1,573617
asked Aug 26 at 23:36
CuriousKid7
1,573617
1,573617
If I don't misunderstand something, shouldn't be the index of the series over the $b_n_i$'s be $i$ and not $n$? Also, interesting question, +1.
– zzuussee
Aug 26 at 23:38
@zzuussee Fixed
– CuriousKid7
Aug 26 at 23:39
2
The sum consisting of all positive terms, $A = sum_i; b_igeq 0 b_i$ converges. Likewise for negative terms, $B = sum_j; b_j < 0 b_j$. Since $A-B = sum left | b_i right |$, $sum left | b_i right |$ must then converge.
– LPenguin
Aug 26 at 23:48
1
@SangchulLee But how do we know the rearrangement preserves the order of the $n_i$?
– CuriousKid7
Aug 26 at 23:49
add a comment |Â
If I don't misunderstand something, shouldn't be the index of the series over the $b_n_i$'s be $i$ and not $n$? Also, interesting question, +1.
– zzuussee
Aug 26 at 23:38
@zzuussee Fixed
– CuriousKid7
Aug 26 at 23:39
2
The sum consisting of all positive terms, $A = sum_i; b_igeq 0 b_i$ converges. Likewise for negative terms, $B = sum_j; b_j < 0 b_j$. Since $A-B = sum left | b_i right |$, $sum left | b_i right |$ must then converge.
– LPenguin
Aug 26 at 23:48
1
@SangchulLee But how do we know the rearrangement preserves the order of the $n_i$?
– CuriousKid7
Aug 26 at 23:49
If I don't misunderstand something, shouldn't be the index of the series over the $b_n_i$'s be $i$ and not $n$? Also, interesting question, +1.
– zzuussee
Aug 26 at 23:38
If I don't misunderstand something, shouldn't be the index of the series over the $b_n_i$'s be $i$ and not $n$? Also, interesting question, +1.
– zzuussee
Aug 26 at 23:38
@zzuussee Fixed
– CuriousKid7
Aug 26 at 23:39
@zzuussee Fixed
– CuriousKid7
Aug 26 at 23:39
2
2
The sum consisting of all positive terms, $A = sum_i; b_igeq 0 b_i$ converges. Likewise for negative terms, $B = sum_j; b_j < 0 b_j$. Since $A-B = sum left | b_i right |$, $sum left | b_i right |$ must then converge.
– LPenguin
Aug 26 at 23:48
The sum consisting of all positive terms, $A = sum_i; b_igeq 0 b_i$ converges. Likewise for negative terms, $B = sum_j; b_j < 0 b_j$. Since $A-B = sum left | b_i right |$, $sum left | b_i right |$ must then converge.
– LPenguin
Aug 26 at 23:48
1
1
@SangchulLee But how do we know the rearrangement preserves the order of the $n_i$?
– CuriousKid7
Aug 26 at 23:49
@SangchulLee But how do we know the rearrangement preserves the order of the $n_i$?
– CuriousKid7
Aug 26 at 23:49
add a comment |Â
1 Answer
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No. Define $$b_n_i=b_n^+=maxb_i,0.$$ By hypothesis, $sum_i=0^infty b_n_i$ converges. Similarly, we can define $b_n^-$ and see that the sum over $b_n^-$ is convergent by hypothesis.
This means that $sum_n=0^infty b_n$ is absolutely convergent since both its positive parts and negative parts converge.
answered Aug 26 at 23:44
Clayton
18.3k22883
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No. Define $$b_n_i=b_n^+=maxb_i,0.$$ By hypothesis, $sum_i=0^infty b_n_i$ converges. Similarly, we can define $b_n^-$ and see that the sum over $b_n^-$ is convergent by hypothesis.
This means that $sum_n=0^infty b_n$ is absolutely convergent since both its positive parts and negative parts converge.
answered Aug 26 at 23:44
Clayton
18.3k22883
add a comment |Â
up vote
2
down vote
accepted
No. Define $$b_n_i=b_n^+=maxb_i,0.$$ By hypothesis, $sum_i=0^infty b_n_i$ converges. Similarly, we can define $b_n^-$ and see that the sum over $b_n^-$ is convergent by hypothesis.
This means that $sum_n=0^infty b_n$ is absolutely convergent since both its positive parts and negative parts converge.
answered Aug 26 at 23:44
Clayton
18.3k22883
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No. Define $$b_n_i=b_n^+=maxb_i,0.$$ By hypothesis, $sum_i=0^infty b_n_i$ converges. Similarly, we can define $b_n^-$ and see that the sum over $b_n^-$ is convergent by hypothesis.
This means that $sum_n=0^infty b_n$ is absolutely convergent since both its positive parts and negative parts converge.
answered Aug 26 at 23:44
Clayton
18.3k22883
No. Define $$b_n_i=b_n^+=maxb_i,0.$$ By hypothesis, $sum_i=0^infty b_n_i$ converges. Similarly, we can define $b_n^-$ and see that the sum over $b_n^-$ is convergent by hypothesis.
This means that $sum_n=0^infty b_n$ is absolutely convergent since both its positive parts and negative parts converge.
answered Aug 26 at 23:44
Clayton
18.3k22883
answered Aug 26 at 23:44
Clayton
18.3k22883
answered Aug 26 at 23:44
Clayton
18.3k22883
answered Aug 26 at 23:44
Clayton
18.3k22883
18.3k22883
add a comment |Â
add a comment |Â
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If I don't misunderstand something, shouldn't be the index of the series over the $b_n_i$'s be $i$ and not $n$? Also, interesting question, +1.
– zzuussee
Aug 26 at 23:38
@zzuussee Fixed
– CuriousKid7
Aug 26 at 23:39
2
The sum consisting of all positive terms, $A = sum_i; b_igeq 0 b_i$ converges. Likewise for negative terms, $B = sum_j; b_j < 0 b_j$. Since $A-B = sum left | b_i right |$, $sum left | b_i right |$ must then converge.
– LPenguin
Aug 26 at 23:48
1
@SangchulLee But how do we know the rearrangement preserves the order of the $n_i$?
– CuriousKid7
Aug 26 at 23:49