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?










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closed as off-topic by Did, amWhy, Shailesh, max_zorn, Rebellos Nov 18 at 11:59


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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?










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    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

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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?










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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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      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.






















          3 Answers
          3






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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.






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          • Nice answer Zwim!!
            – BAYMAX
            Nov 17 at 16:19


















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          1
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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.






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          • 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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          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).






          share|cite|improve this answer




























            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.






            share|cite|improve this answer





















            • Nice answer Zwim!!
              – BAYMAX
              Nov 17 at 16:19















            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.






            share|cite|improve this answer





















            • Nice answer Zwim!!
              – BAYMAX
              Nov 17 at 16:19













            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 17 at 15:53









            zwim

            11.3k628




            11.3k628












            • 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




            Nice answer Zwim!!
            – BAYMAX
            Nov 17 at 16:19










            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.






            share|cite|improve this answer





















            • 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















            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.






            share|cite|improve this answer





















            • 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













            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • 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










            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).






            share|cite|improve this answer

























              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).






              share|cite|improve this answer























                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).






                share|cite|improve this answer












                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).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 13:16









                Bernard

                116k637108




                116k637108















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