Making sure if it is Cauchy











up vote
3
down vote

favorite












In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt frac 1{a^n}$ 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
    1 hour ago















up vote
3
down vote

favorite












In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt frac 1{a^n}$ 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
    1 hour ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt frac 1{a^n}$ 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 $|x_{n+1} - x_n|lt frac 1{a^n}$ 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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









amWhy

191k27223437




191k27223437










asked 1 hour ago









user7857462

311




311












  • slow moving thread -_-
    – Matt A Pelto
    1 hour ago


















  • slow moving thread -_-
    – Matt A Pelto
    1 hour ago
















slow moving thread -_-
– Matt A Pelto
1 hour ago




slow moving thread -_-
– Matt A Pelto
1 hour ago










6 Answers
6






active

oldest

votes

















up vote
3
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





















  • $1/2n$ converges, so it is Cauchy!
    – Bernard
    1 hour ago










  • Yes, I was missing the sum.
    – Michael Lee
    1 hour ago


















up vote
3
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
    1 hour ago










  • 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
    54 mins ago










  • 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
    54 mins ago












  • Oh yeah. My mistake.
    – Chris Custer
    51 mins ago


















up vote
2
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
    1
    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
      1
      down vote













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






      share|cite|improve this answer




























        up vote
        1
        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
          34 mins ago













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














         

        draft saved


        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013628%2fmaking-sure-if-it-is-cauchy%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        6 Answers
        6






        active

        oldest

        votes








        6 Answers
        6






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        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





















        • $1/2n$ converges, so it is Cauchy!
          – Bernard
          1 hour ago










        • Yes, I was missing the sum.
          – Michael Lee
          1 hour ago















        up vote
        3
        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





















        • $1/2n$ converges, so it is Cauchy!
          – Bernard
          1 hour ago










        • Yes, I was missing the sum.
          – Michael Lee
          1 hour ago













        up vote
        3
        down vote










        up vote
        3
        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 1 hour ago









        Michael Lee

        4,6901929




        4,6901929












        • $1/2n$ converges, so it is Cauchy!
          – Bernard
          1 hour ago










        • Yes, I was missing the sum.
          – Michael Lee
          1 hour ago


















        • $1/2n$ converges, so it is Cauchy!
          – Bernard
          1 hour ago










        • Yes, I was missing the sum.
          – Michael Lee
          1 hour ago
















        $1/2n$ converges, so it is Cauchy!
        – Bernard
        1 hour ago




        $1/2n$ converges, so it is Cauchy!
        – Bernard
        1 hour ago












        Yes, I was missing the sum.
        – Michael Lee
        1 hour ago




        Yes, I was missing the sum.
        – Michael Lee
        1 hour ago










        up vote
        3
        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
          1 hour ago










        • 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
          54 mins ago










        • 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
          54 mins ago












        • Oh yeah. My mistake.
          – Chris Custer
          51 mins ago















        up vote
        3
        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
          1 hour ago










        • 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
          54 mins ago










        • 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
          54 mins ago












        • Oh yeah. My mistake.
          – Chris Custer
          51 mins ago













        up vote
        3
        down vote










        up vote
        3
        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 45 mins ago

























        answered 1 hour ago









        Chris Custer

        8,9643623




        8,9643623












        • 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
          1 hour ago










        • 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
          54 mins ago










        • 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
          54 mins ago












        • Oh yeah. My mistake.
          – Chris Custer
          51 mins ago


















        • 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
          1 hour ago










        • 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
          54 mins ago










        • 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
          54 mins ago












        • Oh yeah. My mistake.
          – Chris Custer
          51 mins ago
















        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
        1 hour ago




        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
        1 hour ago












        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
        54 mins ago




        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
        54 mins ago












        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
        54 mins ago






        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
        54 mins ago














        Oh yeah. My mistake.
        – Chris Custer
        51 mins ago




        Oh yeah. My mistake.
        – Chris Custer
        51 mins ago










        up vote
        2
        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
          2
          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
            2
            down vote










            up vote
            2
            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 1 hour ago









            Kavi Rama Murthy

            41.6k31751




            41.6k31751






















                up vote
                1
                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
                  1
                  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
                    1
                    down vote










                    up vote
                    1
                    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 1 hour ago









                    Bernard

                    116k637107




                    116k637107






















                        up vote
                        1
                        down vote













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






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













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






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









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






                            share|cite|improve this answer












                            For $n=1,2,ldots$ define $x_n:=1^{-1}+cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=frac{1}{n+1}$ but the sequence ${x_n}$ is not Cauchy as its terms are just the partial sums of a series that 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










                            answered 1 hour ago









                            Matt A Pelto

                            2,095519




                            2,095519






















                                up vote
                                1
                                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
                                  34 mins ago

















                                up vote
                                1
                                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
                                  34 mins ago















                                up vote
                                1
                                down vote










                                up vote
                                1
                                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 1 hour ago









                                Trevor Gunn

                                13.8k32045




                                13.8k32045








                                • 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
                                  34 mins ago
















                                • 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
                                  34 mins ago










                                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
                                34 mins ago






                                +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
                                34 mins ago




















                                 

                                draft saved


                                draft discarded



















































                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013628%2fmaking-sure-if-it-is-cauchy%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Futebolista

                                Lallio

                                Jornalista