Making sure if it is Cauchy












6














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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  • 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


















6














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







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
















6












6








6


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?






real-analysis cauchy-sequences







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




real-analysis cauchy-sequences






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 5:41









Matt A Pelto

2,368620




2,368620










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
6






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oldest

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11














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



















8














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





























    6














    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





























      4














      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





















      3














      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































        0














        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





















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          6 Answers
          6






          active

          oldest

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          6 Answers
          6






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          11














          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
















          11














          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














          11












          11








          11






          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

          10.6k3724




          10.6k3724












          • 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











          8














          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


























            8














            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
























              8












              8








              8






              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























                  6














                  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


























                    6














                    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
























                      6












                      6








                      6






                      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

                      48.9k31854




                      48.9k31854























                          4














                          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


















                          4














                          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
















                          4












                          4








                          4






                          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

                          14.1k32046




                          14.1k32046








                          • 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













                          3














                          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




























                            3














                            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


























                              3












                              3








                              3






                              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

                              2,368620




                              2,368620























                                  0














                                  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


























                                    0














                                    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
























                                      0












                                      0








                                      0






                                      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

                                      118k638111




                                      118k638111






























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