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












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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.






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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.




















    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













    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.


















      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













      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













      share|cite|improve this question











      share|cite|improve this question




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      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.
























          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.)






          share|cite|improve this answer




























            1 Answer
            1






            active

            oldest

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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


























              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
























                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












                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












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 '18 at 11:23









                p4sch

                4,760217




                4,760217















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