$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






limits convergence







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






limits convergence







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






limits convergence







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




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

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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











      $endgroup$













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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






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        active

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        active

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        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






























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