$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n}...












1












$begingroup$


Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.






My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:




  • Is the statement really true?

  • (If so, how could you prove it?)


Cheers,

Pascal










share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    $$ a_n = n.{} $$
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 22:08






  • 1




    $begingroup$
    @MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
    $endgroup$
    – plauer
    Nov 26 '18 at 22:11
















1












$begingroup$


Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.






My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:




  • Is the statement really true?

  • (If so, how could you prove it?)


Cheers,

Pascal










share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    $$ a_n = n.{} $$
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 22:08






  • 1




    $begingroup$
    @MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
    $endgroup$
    – plauer
    Nov 26 '18 at 22:11














1












1








1





$begingroup$


Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.






My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:




  • Is the statement really true?

  • (If so, how could you prove it?)


Cheers,

Pascal










share|cite|improve this question









$endgroup$




Let $(a_n)_{nin mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1 Rightarrow exists c in mathbb{R}: limlimits_{n to infty} {a_n} = c$$
Put into words: If $limlimits_{n to infty} frac{a_n}{a_{n+1}} = 1$ then $limlimits_{n to infty} a_{n}$ converges.






My intuition behind $lim_{n to infty} frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{nin mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:




  • Is the statement really true?

  • (If so, how could you prove it?)


Cheers,

Pascal







limits convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 '18 at 22:07









plauerplauer

305




305








  • 3




    $begingroup$
    $$ a_n = n.{} $$
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 22:08






  • 1




    $begingroup$
    @MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
    $endgroup$
    – plauer
    Nov 26 '18 at 22:11














  • 3




    $begingroup$
    $$ a_n = n.{} $$
    $endgroup$
    – MisterRiemann
    Nov 26 '18 at 22:08






  • 1




    $begingroup$
    @MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
    $endgroup$
    – plauer
    Nov 26 '18 at 22:11








3




3




$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08




$begingroup$
$$ a_n = n.{} $$
$endgroup$
– MisterRiemann
Nov 26 '18 at 22:08




1




1




$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11




$begingroup$
@MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then.
$endgroup$
– plauer
Nov 26 '18 at 22:11










3 Answers
3






active

oldest

votes


















4












$begingroup$

As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.



Therefore unfortunately your guess is definitely not true!





As a remark, other common myths on limits are:



1) $a_n to infty implies a_{n+1}ge a_n$



2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$



3) $a_n to 0^+ implies a_{n+1}le a_n$



4) $a_n$ bounded $implies a_n to L$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Thanks to MisterRiemann for the counterexample of $a_n = n$.



    $$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
    but:
    $$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      The conclusion is wrong even for bounded sequences.



      An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
      $$
      leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
      = leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
      le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
      = leftlvert frac{cos(log(x_n))}{x_n}rightrvert
      $$

      for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
      $$
      leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
      $$

      i.e. $frac{a_{n+1}}{a_n} to 1$.



      Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
      so that the ratio of successive sequence elements approaches one.






      share|cite|improve this answer











      $endgroup$













        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',
        autoActivateHeartbeat: false,
        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%2f3015009%2flim-limits-n-to-infty-fraca-na-n1-1-rightarrow-exists-c-in-m%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.



        Therefore unfortunately your guess is definitely not true!





        As a remark, other common myths on limits are:



        1) $a_n to infty implies a_{n+1}ge a_n$



        2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$



        3) $a_n to 0^+ implies a_{n+1}le a_n$



        4) $a_n$ bounded $implies a_n to L$






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.



          Therefore unfortunately your guess is definitely not true!





          As a remark, other common myths on limits are:



          1) $a_n to infty implies a_{n+1}ge a_n$



          2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$



          3) $a_n to 0^+ implies a_{n+1}le a_n$



          4) $a_n$ bounded $implies a_n to L$






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.



            Therefore unfortunately your guess is definitely not true!





            As a remark, other common myths on limits are:



            1) $a_n to infty implies a_{n+1}ge a_n$



            2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$



            3) $a_n to 0^+ implies a_{n+1}le a_n$



            4) $a_n$ bounded $implies a_n to L$






            share|cite|improve this answer











            $endgroup$



            As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=log n$ and so on.



            Therefore unfortunately your guess is definitely not true!





            As a remark, other common myths on limits are:



            1) $a_n to infty implies a_{n+1}ge a_n$



            2) $a_n to L implies a_n to L^+ quad lor quad a_n to L^-$



            3) $a_n to 0^+ implies a_{n+1}le a_n$



            4) $a_n$ bounded $implies a_n to L$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 26 '18 at 22:34

























            answered Nov 26 '18 at 22:12









            gimusigimusi

            92.8k84494




            92.8k84494























                1












                $begingroup$

                Thanks to MisterRiemann for the counterexample of $a_n = n$.



                $$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
                but:
                $$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Thanks to MisterRiemann for the counterexample of $a_n = n$.



                  $$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
                  but:
                  $$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Thanks to MisterRiemann for the counterexample of $a_n = n$.



                    $$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
                    but:
                    $$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$






                    share|cite|improve this answer









                    $endgroup$



                    Thanks to MisterRiemann for the counterexample of $a_n = n$.



                    $$limlimits_{n to infty} frac{a_n}{a_{n+1}} = limlimits_{n to infty} frac{n}{{n+1}} = 1$$
                    but:
                    $$limlimits_{n to infty} {a_n} =limlimits_{n to infty} n = infty Rightarrow nexists c in mathbb{R}: limlimits_{n to infty} {a_n} = c $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 26 '18 at 22:19









                    plauerplauer

                    305




                    305























                        1












                        $begingroup$

                        The conclusion is wrong even for bounded sequences.



                        An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
                        $$
                        leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
                        = leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
                        le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
                        = leftlvert frac{cos(log(x_n))}{x_n}rightrvert
                        $$

                        for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
                        $$
                        leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
                        $$

                        i.e. $frac{a_{n+1}}{a_n} to 1$.



                        Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
                        so that the ratio of successive sequence elements approaches one.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          The conclusion is wrong even for bounded sequences.



                          An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
                          $$
                          leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
                          = leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
                          le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
                          = leftlvert frac{cos(log(x_n))}{x_n}rightrvert
                          $$

                          for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
                          $$
                          leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
                          $$

                          i.e. $frac{a_{n+1}}{a_n} to 1$.



                          Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
                          so that the ratio of successive sequence elements approaches one.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            The conclusion is wrong even for bounded sequences.



                            An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
                            $$
                            leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
                            = leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
                            le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
                            = leftlvert frac{cos(log(x_n))}{x_n}rightrvert
                            $$

                            for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
                            $$
                            leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
                            $$

                            i.e. $frac{a_{n+1}}{a_n} to 1$.



                            Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
                            so that the ratio of successive sequence elements approaches one.






                            share|cite|improve this answer











                            $endgroup$



                            The conclusion is wrong even for bounded sequences.



                            An example is $a_ n = 2 + sin(log(n))$ which is not convergent (it oscillates between $1$ and $3$). But
                            $$
                            leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert
                            = leftlvert frac{sin(log(n+1))- sin(log(n))}{2 + sin(log(n))} rightrvert \
                            le leftlvert sin(log(n+1))- sin(log(n)) rightrvert
                            = leftlvert frac{cos(log(x_n))}{x_n}rightrvert
                            $$

                            for some $x_n in (n, n+1)$, using the mean value theorem. It follows that
                            $$
                            leftlvert frac{a_{n+1}}{a_n} - 1 rightrvert le frac{1}{n} to 0 , ,
                            $$

                            i.e. $frac{a_{n+1}}{a_n} to 1$.



                            Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency,
                            so that the ratio of successive sequence elements approaches one.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 27 '18 at 13:18

























                            answered Nov 27 '18 at 12:58









                            Martin RMartin R

                            27.9k33255




                            27.9k33255






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015009%2flim-limits-n-to-infty-fraca-na-n1-1-rightarrow-exists-c-in-m%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

                                How to change which sound is reproduced for terminal bell?

                                Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                                Can I use Tabulator js library in my java Spring + Thymeleaf project?