Can we make a convergent series absolutely convergent under certain conditions? [closed]

2

Suppose $sum a_n$ is convergent,can we impose some condition on this series to make the series absolutely convergent?

Except some trivial conditions that it has finitely many positive or negative terms etc.

sequences-and-series

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asked Nov 20 '18 at 2:41

Tom.

14118

closed as unclear what you're asking by Kavi Rama Murthy, mau, José Carlos Santos, Christopher, Zvi Nov 20 '18 at 12:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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2

Suppose $sum a_n$ is convergent,can we impose some condition on this series to make the series absolutely convergent?

Except some trivial conditions that it has finitely many positive or negative terms etc.

sequences-and-series

share|cite|improve this question

asked Nov 20 '18 at 2:41

Tom.

14118

closed as unclear what you're asking by Kavi Rama Murthy, mau, José Carlos Santos, Christopher, Zvi Nov 20 '18 at 12:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

2

2

2

Suppose $sum a_n$ is convergent,can we impose some condition on this series to make the series absolutely convergent?

Except some trivial conditions that it has finitely many positive or negative terms etc.

sequences-and-series

share|cite|improve this question

asked Nov 20 '18 at 2:41

Tom.

14118

Suppose $sum a_n$ is convergent,can we impose some condition on this series to make the series absolutely convergent?

Except some trivial conditions that it has finitely many positive or negative terms etc.

sequences-and-series

sequences-and-series

share|cite|improve this question

asked Nov 20 '18 at 2:41

Tom.

14118

share|cite|improve this question

asked Nov 20 '18 at 2:41

Tom.

14118

share|cite|improve this question

share|cite|improve this question

asked Nov 20 '18 at 2:41

Tom.

14118

asked Nov 20 '18 at 2:41

Tom.

14118

asked Nov 20 '18 at 2:41

Tom.

14118

14118

closed as unclear what you're asking by Kavi Rama Murthy, mau, José Carlos Santos, Christopher, Zvi Nov 20 '18 at 12:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

closed as unclear what you're asking by Kavi Rama Murthy, mau, José Carlos Santos, Christopher, Zvi Nov 20 '18 at 12:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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1 Answer
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In $mathbb{R}$ (resp. $mathbb{C}$) the absolute convergence is equivalent to the condition that for any permutation $pi colon mathbb{N} rightarrow mathbb{N}$ the series $sum_{n=1}^infty a_{pi(n)}$ is convergent (known as Riemann rearrangement theorem). In general, you cannot say more! (Of course, you could use e.g. the integral comparison test, if $|a_n|$ can be extenend to an monotically decreasing function $f$ in order to get absolute convergence. However, this is - of course - only a very special case.)

Note that by the Theorem of Dvoretzky-Rogers this statement is false in infinite dimensional banach spaces. (There is an unconditional convergent series which is not absolute convergent.)

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answered Nov 20 '18 at 11:23

p4sch

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1 Answer
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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

In $mathbb{R}$ (resp. $mathbb{C}$) the absolute convergence is equivalent to the condition that for any permutation $pi colon mathbb{N} rightarrow mathbb{N}$ the series $sum_{n=1}^infty a_{pi(n)}$ is convergent (known as Riemann rearrangement theorem). In general, you cannot say more! (Of course, you could use e.g. the integral comparison test, if $|a_n|$ can be extenend to an monotically decreasing function $f$ in order to get absolute convergence. However, this is - of course - only a very special case.)

Note that by the Theorem of Dvoretzky-Rogers this statement is false in infinite dimensional banach spaces. (There is an unconditional convergent series which is not absolute convergent.)

share|cite|improve this answer

answered Nov 20 '18 at 11:23

p4sch

4,760217

add a comment | 

0

In $mathbb{R}$ (resp. $mathbb{C}$) the absolute convergence is equivalent to the condition that for any permutation $pi colon mathbb{N} rightarrow mathbb{N}$ the series $sum_{n=1}^infty a_{pi(n)}$ is convergent (known as Riemann rearrangement theorem). In general, you cannot say more! (Of course, you could use e.g. the integral comparison test, if $|a_n|$ can be extenend to an monotically decreasing function $f$ in order to get absolute convergence. However, this is - of course - only a very special case.)

Note that by the Theorem of Dvoretzky-Rogers this statement is false in infinite dimensional banach spaces. (There is an unconditional convergent series which is not absolute convergent.)

share|cite|improve this answer

answered Nov 20 '18 at 11:23

p4sch

4,760217

add a comment | 

0

0

0

In $mathbb{R}$ (resp. $mathbb{C}$) the absolute convergence is equivalent to the condition that for any permutation $pi colon mathbb{N} rightarrow mathbb{N}$ the series $sum_{n=1}^infty a_{pi(n)}$ is convergent (known as Riemann rearrangement theorem). In general, you cannot say more! (Of course, you could use e.g. the integral comparison test, if $|a_n|$ can be extenend to an monotically decreasing function $f$ in order to get absolute convergence. However, this is - of course - only a very special case.)

Note that by the Theorem of Dvoretzky-Rogers this statement is false in infinite dimensional banach spaces. (There is an unconditional convergent series which is not absolute convergent.)

share|cite|improve this answer

answered Nov 20 '18 at 11:23

p4sch

4,760217

In $mathbb{R}$ (resp. $mathbb{C}$) the absolute convergence is equivalent to the condition that for any permutation $pi colon mathbb{N} rightarrow mathbb{N}$ the series $sum_{n=1}^infty a_{pi(n)}$ is convergent (known as Riemann rearrangement theorem). In general, you cannot say more! (Of course, you could use e.g. the integral comparison test, if $|a_n|$ can be extenend to an monotically decreasing function $f$ in order to get absolute convergence. However, this is - of course - only a very special case.)

Note that by the Theorem of Dvoretzky-Rogers this statement is false in infinite dimensional banach spaces. (There is an unconditional convergent series which is not absolute convergent.)

share|cite|improve this answer

answered Nov 20 '18 at 11:23

p4sch

4,760217

share|cite|improve this answer

share|cite|improve this answer

answered Nov 20 '18 at 11:23

p4sch

4,760217

answered Nov 20 '18 at 11:23

p4sch

4,760217

answered Nov 20 '18 at 11:23

p4sch

4,760217

4,760217

add a comment | 

add a comment | 

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