Making sure if it is Cauchy











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In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.



This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?










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  • slow moving thread -_-
    – Matt A Pelto
    Nov 26 at 0:38










  • Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
    – Dirk
    Nov 26 at 5:31










  • Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
    – Matt A Pelto
    Nov 26 at 5:44

















up vote
6
down vote

favorite
1












In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.



This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?










share|cite|improve this question
























  • slow moving thread -_-
    – Matt A Pelto
    Nov 26 at 0:38










  • Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
    – Dirk
    Nov 26 at 5:31










  • Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
    – Matt A Pelto
    Nov 26 at 5:44















up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.



This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?










share|cite|improve this question















In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ then $(x_n)$ is a Cauchy sequence.



This was solved successfully but the question is if $|x_{n+1} - x_n|lt frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?







real-analysis cauchy-sequences






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edited Nov 26 at 5:41









Matt A Pelto

2,338620




2,338620










asked Nov 26 at 0:29









user7857462

463




463












  • slow moving thread -_-
    – Matt A Pelto
    Nov 26 at 0:38










  • Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
    – Dirk
    Nov 26 at 5:31










  • Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
    – Matt A Pelto
    Nov 26 at 5:44




















  • slow moving thread -_-
    – Matt A Pelto
    Nov 26 at 0:38










  • Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
    – Dirk
    Nov 26 at 5:31










  • Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
    – Matt A Pelto
    Nov 26 at 5:44


















slow moving thread -_-
– Matt A Pelto
Nov 26 at 0:38




slow moving thread -_-
– Matt A Pelto
Nov 26 at 0:38












Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
– Dirk
Nov 26 at 5:31




Are you about the first question? I am confused the the "for all a" and even more by the "a<1" (which makes the upper bound on the differences rather large).
– Dirk
Nov 26 at 5:31












Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
– Matt A Pelto
Nov 26 at 5:44






Seems to be some sort of error. The proper correction should be either changing "$frac 1{a^n}$" to "$a^n$" or changing "$a<1$" to "$a>1$" (minding the radius of convergence for geometric series).
– Matt A Pelto
Nov 26 at 5:44












6 Answers
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11
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Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.



$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.



But it diverges.






share|cite|improve this answer























  • This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
    – Michael Lee
    Nov 26 at 0:55










  • Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
    – Chris Custer
    Nov 26 at 1:08










  • The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
    – Michael Lee
    Nov 26 at 1:08












  • Oh yeah. My mistake.
    – Chris Custer
    Nov 26 at 1:10


















up vote
8
down vote













No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.






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    up vote
    6
    down vote













    Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.






    share|cite|improve this answer




























      up vote
      4
      down vote













      This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,



      begin{align}
      |x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
      &le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
      &le a_n + a_{n + 1} + dots + a_{n + m - 1} \
      &le sum_{k = n}^infty a_k
      end{align}



      Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,



      $$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$



      Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,



      $$ |x_n - x_{n + m}| < varepsilon $$



      Which means the sequence $(x_n)$ is Cauchy.



      If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.






      share|cite|improve this answer

















      • 1




        +1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
        – Matt A Pelto
        Nov 26 at 1:28




















      up vote
      3
      down vote













      For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.






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        This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).






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          6 Answers
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          6 Answers
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          up vote
          11
          down vote













          Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.



          $mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.



          But it diverges.






          share|cite|improve this answer























          • This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
            – Michael Lee
            Nov 26 at 0:55










          • Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
            – Chris Custer
            Nov 26 at 1:08










          • The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
            – Michael Lee
            Nov 26 at 1:08












          • Oh yeah. My mistake.
            – Chris Custer
            Nov 26 at 1:10















          up vote
          11
          down vote













          Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.



          $mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.



          But it diverges.






          share|cite|improve this answer























          • This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
            – Michael Lee
            Nov 26 at 0:55










          • Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
            – Chris Custer
            Nov 26 at 1:08










          • The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
            – Michael Lee
            Nov 26 at 1:08












          • Oh yeah. My mistake.
            – Chris Custer
            Nov 26 at 1:10













          up vote
          11
          down vote










          up vote
          11
          down vote









          Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.



          $mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.



          But it diverges.






          share|cite|improve this answer














          Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.



          $mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.



          But it diverges.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 at 1:17

























          answered Nov 26 at 0:37









          Chris Custer

          9,7743624




          9,7743624












          • This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
            – Michael Lee
            Nov 26 at 0:55










          • Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
            – Chris Custer
            Nov 26 at 1:08










          • The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
            – Michael Lee
            Nov 26 at 1:08












          • Oh yeah. My mistake.
            – Chris Custer
            Nov 26 at 1:10


















          • This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
            – Michael Lee
            Nov 26 at 0:55










          • Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
            – Chris Custer
            Nov 26 at 1:08










          • The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
            – Michael Lee
            Nov 26 at 1:08












          • Oh yeah. My mistake.
            – Chris Custer
            Nov 26 at 1:10
















          This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
          – Michael Lee
          Nov 26 at 0:55




          This is the difference between $1/n$ and $1/(n+1)$, not between two terms of the sequence of partial sums. The sequence $(1/n)_{nin mathbb{N}}$ is indeed Cauchy.
          – Michael Lee
          Nov 26 at 0:55












          Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
          – Chris Custer
          Nov 26 at 1:08




          Yes. I'm referring to the sequence of partial sums of the series. It's not Cauchy because it doesn't converge. I was addressing the OP's example. This shows that $mid a_{n+1}-a_nmid$ can be less than $frac1n$, but the sequence can still not be Cauchy.
          – Chris Custer
          Nov 26 at 1:08












          The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
          – Michael Lee
          Nov 26 at 1:08






          The difference between partial sums is $1/(n+1)$, not $1/n(n+1)$. You add $1/(n+1)$ to get from the $n$th partial sum to the $(n+1)$th.
          – Michael Lee
          Nov 26 at 1:08














          Oh yeah. My mistake.
          – Chris Custer
          Nov 26 at 1:10




          Oh yeah. My mistake.
          – Chris Custer
          Nov 26 at 1:10










          up vote
          8
          down vote













          No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.






          share|cite|improve this answer

























            up vote
            8
            down vote













            No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.






            share|cite|improve this answer























              up vote
              8
              down vote










              up vote
              8
              down vote









              No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.






              share|cite|improve this answer












              No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 26 at 0:34









              Michael Lee

              4,7401929




              4,7401929






















                  up vote
                  6
                  down vote













                  Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.






                  share|cite|improve this answer

























                    up vote
                    6
                    down vote













                    Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.






                    share|cite|improve this answer























                      up vote
                      6
                      down vote










                      up vote
                      6
                      down vote









                      Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.






                      share|cite|improve this answer












                      Take $x_n=1+frac 1 2+cdots+frac 1 n$. This is not Cauchy because the harmonic series $1+frac 1 2+cdots$ is divergent.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 26 at 0:34









                      Kavi Rama Murthy

                      45.4k31853




                      45.4k31853






















                          up vote
                          4
                          down vote













                          This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,



                          begin{align}
                          |x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
                          &le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
                          &le a_n + a_{n + 1} + dots + a_{n + m - 1} \
                          &le sum_{k = n}^infty a_k
                          end{align}



                          Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,



                          $$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$



                          Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,



                          $$ |x_n - x_{n + m}| < varepsilon $$



                          Which means the sequence $(x_n)$ is Cauchy.



                          If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.






                          share|cite|improve this answer

















                          • 1




                            +1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
                            – Matt A Pelto
                            Nov 26 at 1:28

















                          up vote
                          4
                          down vote













                          This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,



                          begin{align}
                          |x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
                          &le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
                          &le a_n + a_{n + 1} + dots + a_{n + m - 1} \
                          &le sum_{k = n}^infty a_k
                          end{align}



                          Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,



                          $$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$



                          Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,



                          $$ |x_n - x_{n + m}| < varepsilon $$



                          Which means the sequence $(x_n)$ is Cauchy.



                          If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.






                          share|cite|improve this answer

















                          • 1




                            +1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
                            – Matt A Pelto
                            Nov 26 at 1:28















                          up vote
                          4
                          down vote










                          up vote
                          4
                          down vote









                          This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,



                          begin{align}
                          |x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
                          &le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
                          &le a_n + a_{n + 1} + dots + a_{n + m - 1} \
                          &le sum_{k = n}^infty a_k
                          end{align}



                          Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,



                          $$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$



                          Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,



                          $$ |x_n - x_{n + m}| < varepsilon $$



                          Which means the sequence $(x_n)$ is Cauchy.



                          If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.






                          share|cite|improve this answer












                          This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $sum_{k = 0}^infty a_k < infty$ because, if this condition holds,



                          begin{align}
                          |x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - cdots + x_{n + m - 1} - x_{n + m}| \
                          &le |x_{n} - x_{n + 1}| + cdots + |x_{n + m - 1} - x_{n + m}| \
                          &le a_n + a_{n + 1} + dots + a_{n + m - 1} \
                          &le sum_{k = n}^infty a_k
                          end{align}



                          Now convergence of $sum a_k$ to $A$ means that for any $varepsilon > 0$ there exists $N$ such that for all $n ge N$,



                          $$ left| A - sum_{k = 0}^{n - 1} a_k right| = sum_{k = n}^infty a_k < varepsilon $$



                          Comparing this with the above, we have for every $n ge N$ and every $m ge 0$,



                          $$ |x_n - x_{n + m}| < varepsilon $$



                          Which means the sequence $(x_n)$ is Cauchy.



                          If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 26 at 0:46









                          Trevor Gunn

                          14k32046




                          14k32046








                          • 1




                            +1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
                            – Matt A Pelto
                            Nov 26 at 1:28
















                          • 1




                            +1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
                            – Matt A Pelto
                            Nov 26 at 1:28










                          1




                          1




                          +1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
                          – Matt A Pelto
                          Nov 26 at 1:28






                          +1 for the most informative response and using varepsilon...no effort has been spared here. I will however nitpick at the last sentence by noting that a different trick may or may not still work, depending on whether the sequence is indeed Cauchy which I presume you know but just add as clarification for people such as the question asker. An example where a different trick might apply is the sequence of partial sums of the alternating series $sum_{n=1}^infty frac{(-1)^{n+1}}{n}$.
                          – Matt A Pelto
                          Nov 26 at 1:28












                          up vote
                          3
                          down vote













                          For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.






                          share|cite|improve this answer



























                            up vote
                            3
                            down vote













                            For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.






                            share|cite|improve this answer

























                              up vote
                              3
                              down vote










                              up vote
                              3
                              down vote









                              For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.






                              share|cite|improve this answer














                              For each $n in mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}_{n=1}^infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $mathbb{R}$ is complete.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 30 at 3:28

























                              answered Nov 26 at 0:38









                              Matt A Pelto

                              2,338620




                              2,338620






















                                  up vote
                                  0
                                  down vote













                                  This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).






                                      share|cite|improve this answer












                                      This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 26 at 0:37









                                      Bernard

                                      117k637109




                                      117k637109






























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