Two-sided logarithm inequality












3












$begingroup$


I couldn't find a duplicate question, so I apologize if this has been asked before.



I'm trying to show that



$$ n - 1 < left(log left( frac{n}{n-1}right)right)^{-1} < n tag{1} $$



I've verified this numerically, and it even seems to be the case that



$$ lim_{n to infty} frac{1}{log left( n / (n - 1)right)} = n - frac{1}{2} $$



Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then



$$ log(n) - log(n - 1) sim frac{1}{n} tag{2} $$



However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.



Hints are definitely welcome.






limits inequality logarithms







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Cf. this question
    $endgroup$
    – J. W. Tanner
    2 days ago
















3












$begingroup$


I couldn't find a duplicate question, so I apologize if this has been asked before.



I'm trying to show that



$$ n - 1 < left(log left( frac{n}{n-1}right)right)^{-1} < n tag{1} $$



I've verified this numerically, and it even seems to be the case that



$$ lim_{n to infty} frac{1}{log left( n / (n - 1)right)} = n - frac{1}{2} $$



Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then



$$ log(n) - log(n - 1) sim frac{1}{n} tag{2} $$



However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.



Hints are definitely welcome.






limits inequality logarithms







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Cf. this question
    $endgroup$
    – J. W. Tanner
    2 days ago














3












3








3





$begingroup$


I couldn't find a duplicate question, so I apologize if this has been asked before.



I'm trying to show that



$$ n - 1 < left(log left( frac{n}{n-1}right)right)^{-1} < n tag{1} $$



I've verified this numerically, and it even seems to be the case that



$$ lim_{n to infty} frac{1}{log left( n / (n - 1)right)} = n - frac{1}{2} $$



Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then



$$ log(n) - log(n - 1) sim frac{1}{n} tag{2} $$



However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.



Hints are definitely welcome.






limits inequality logarithms







share|cite|improve this question











$endgroup$




I couldn't find a duplicate question, so I apologize if this has been asked before.



I'm trying to show that



$$ n - 1 < left(log left( frac{n}{n-1}right)right)^{-1} < n tag{1} $$



I've verified this numerically, and it even seems to be the case that



$$ lim_{n to infty} frac{1}{log left( n / (n - 1)right)} = n - frac{1}{2} $$



Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then



$$ log(n) - log(n - 1) sim frac{1}{n} tag{2} $$



However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.



Hints are definitely welcome.






limits inequality logarithms




limits inequality logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









egreg

185k1486206




185k1486206










asked 2 days ago









Enrico BorbaEnrico Borba

451139




451139








  • 3




    $begingroup$
    Cf. this question
    $endgroup$
    – J. W. Tanner
    2 days ago














  • 3




    $begingroup$
    Cf. this question
    $endgroup$
    – J. W. Tanner
    2 days ago








3




3




$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
2 days ago




$begingroup$
Cf. this question
$endgroup$
– J. W. Tanner
2 days ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

Your inequality is equivalent to$$frac1{n-1}>logleft(frac n{n-1}right)>frac1n,$$which, in turn, is equivalent to$$frac1{n-1}>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_{n-1}^nfrac{mathrm dt}t.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Awesome hint. Thanks!
    $endgroup$
    – Enrico Borba
    2 days ago










  • $begingroup$
    I'm glad I could help.
    $endgroup$
    – José Carlos Santos
    2 days ago



















0












$begingroup$

Let's try with a reductio ad absurdum :



1 disequality



Suppose that for some $n$:



$$log^{-1}(frac{n}{n-1})<n-1 $$



Notice that $log^{-1}(frac{n}{n-1})=log_{frac{n}{n-1}}(e)$ so:



$log_{frac{n}{n-1}}(e)<n-1$



$(frac{n}{n-1})^{n-1}>e$



Now $ n-1=x $:



$(1+frac{1}{x})^{x}>e$



But this is absurd because $(1+frac{1}{x})^{x}$ is strictly increasing and his limit is $e$.



2 disequality
As before:
$$log^{-1}(frac{n}{n-1})>n $$



$log_{frac{n}{n-1}}(e)>n$



$(frac{n}{n-1})^{n}<e$



And this is absurd because that function is strictly decreasing and his limit value is $e$






share|cite|improve this answer








New contributor




Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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    2 Answers
    2






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    2 Answers
    2






    active

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    active

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    7












    $begingroup$

    Your inequality is equivalent to$$frac1{n-1}>logleft(frac n{n-1}right)>frac1n,$$which, in turn, is equivalent to$$frac1{n-1}>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_{n-1}^nfrac{mathrm dt}t.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Awesome hint. Thanks!
      $endgroup$
      – Enrico Borba
      2 days ago










    • $begingroup$
      I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      2 days ago
















    7












    $begingroup$

    Your inequality is equivalent to$$frac1{n-1}>logleft(frac n{n-1}right)>frac1n,$$which, in turn, is equivalent to$$frac1{n-1}>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_{n-1}^nfrac{mathrm dt}t.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Awesome hint. Thanks!
      $endgroup$
      – Enrico Borba
      2 days ago










    • $begingroup$
      I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      2 days ago














    7












    7








    7





    $begingroup$

    Your inequality is equivalent to$$frac1{n-1}>logleft(frac n{n-1}right)>frac1n,$$which, in turn, is equivalent to$$frac1{n-1}>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_{n-1}^nfrac{mathrm dt}t.$$






    share|cite|improve this answer











    $endgroup$



    Your inequality is equivalent to$$frac1{n-1}>logleft(frac n{n-1}right)>frac1n,$$which, in turn, is equivalent to$$frac1{n-1}>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_{n-1}^nfrac{mathrm dt}t.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    José Carlos SantosJosé Carlos Santos

    170k23132238




    170k23132238












    • $begingroup$
      Awesome hint. Thanks!
      $endgroup$
      – Enrico Borba
      2 days ago










    • $begingroup$
      I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      2 days ago


















    • $begingroup$
      Awesome hint. Thanks!
      $endgroup$
      – Enrico Borba
      2 days ago










    • $begingroup$
      I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      2 days ago
















    $begingroup$
    Awesome hint. Thanks!
    $endgroup$
    – Enrico Borba
    2 days ago




    $begingroup$
    Awesome hint. Thanks!
    $endgroup$
    – Enrico Borba
    2 days ago












    $begingroup$
    I'm glad I could help.
    $endgroup$
    – José Carlos Santos
    2 days ago




    $begingroup$
    I'm glad I could help.
    $endgroup$
    – José Carlos Santos
    2 days ago











    0












    $begingroup$

    Let's try with a reductio ad absurdum :



    1 disequality



    Suppose that for some $n$:



    $$log^{-1}(frac{n}{n-1})<n-1 $$



    Notice that $log^{-1}(frac{n}{n-1})=log_{frac{n}{n-1}}(e)$ so:



    $log_{frac{n}{n-1}}(e)<n-1$



    $(frac{n}{n-1})^{n-1}>e$



    Now $ n-1=x $:



    $(1+frac{1}{x})^{x}>e$



    But this is absurd because $(1+frac{1}{x})^{x}$ is strictly increasing and his limit is $e$.



    2 disequality
    As before:
    $$log^{-1}(frac{n}{n-1})>n $$



    $log_{frac{n}{n-1}}(e)>n$



    $(frac{n}{n-1})^{n}<e$



    And this is absurd because that function is strictly decreasing and his limit value is $e$






    share|cite|improve this answer








    New contributor




    Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$


















      0












      $begingroup$

      Let's try with a reductio ad absurdum :



      1 disequality



      Suppose that for some $n$:



      $$log^{-1}(frac{n}{n-1})<n-1 $$



      Notice that $log^{-1}(frac{n}{n-1})=log_{frac{n}{n-1}}(e)$ so:



      $log_{frac{n}{n-1}}(e)<n-1$



      $(frac{n}{n-1})^{n-1}>e$



      Now $ n-1=x $:



      $(1+frac{1}{x})^{x}>e$



      But this is absurd because $(1+frac{1}{x})^{x}$ is strictly increasing and his limit is $e$.



      2 disequality
      As before:
      $$log^{-1}(frac{n}{n-1})>n $$



      $log_{frac{n}{n-1}}(e)>n$



      $(frac{n}{n-1})^{n}<e$



      And this is absurd because that function is strictly decreasing and his limit value is $e$






      share|cite|improve this answer








      New contributor




      Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$
















        0












        0








        0





        $begingroup$

        Let's try with a reductio ad absurdum :



        1 disequality



        Suppose that for some $n$:



        $$log^{-1}(frac{n}{n-1})<n-1 $$



        Notice that $log^{-1}(frac{n}{n-1})=log_{frac{n}{n-1}}(e)$ so:



        $log_{frac{n}{n-1}}(e)<n-1$



        $(frac{n}{n-1})^{n-1}>e$



        Now $ n-1=x $:



        $(1+frac{1}{x})^{x}>e$



        But this is absurd because $(1+frac{1}{x})^{x}$ is strictly increasing and his limit is $e$.



        2 disequality
        As before:
        $$log^{-1}(frac{n}{n-1})>n $$



        $log_{frac{n}{n-1}}(e)>n$



        $(frac{n}{n-1})^{n}<e$



        And this is absurd because that function is strictly decreasing and his limit value is $e$






        share|cite|improve this answer








        New contributor




        Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$



        Let's try with a reductio ad absurdum :



        1 disequality



        Suppose that for some $n$:



        $$log^{-1}(frac{n}{n-1})<n-1 $$



        Notice that $log^{-1}(frac{n}{n-1})=log_{frac{n}{n-1}}(e)$ so:



        $log_{frac{n}{n-1}}(e)<n-1$



        $(frac{n}{n-1})^{n-1}>e$



        Now $ n-1=x $:



        $(1+frac{1}{x})^{x}>e$



        But this is absurd because $(1+frac{1}{x})^{x}$ is strictly increasing and his limit is $e$.



        2 disequality
        As before:
        $$log^{-1}(frac{n}{n-1})>n $$



        $log_{frac{n}{n-1}}(e)>n$



        $(frac{n}{n-1})^{n}<e$



        And this is absurd because that function is strictly decreasing and his limit value is $e$







        share|cite|improve this answer








        New contributor




        Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer






        New contributor




        Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 2 days ago









        EurekaEureka

        21611




        21611




        New contributor




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        New contributor





        Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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