Can someone solve this question: Show that $ln(1+n)<n$ for all $nge1$ [closed]












0












$begingroup$


I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.



I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?










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closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 6




    $begingroup$
    Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
    $endgroup$
    – RRL
    Jan 9 at 6:03






  • 1




    $begingroup$
    just try first few integers, and make a plot to see it. It will help you to get the idea
    $endgroup$
    – Lee
    Jan 9 at 6:15






  • 5




    $begingroup$
    See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:16






  • 1




    $begingroup$
    FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:53








  • 1




    $begingroup$
    You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
    $endgroup$
    – Zubzub
    Jan 9 at 7:48
















0












$begingroup$


I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.



I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?










share|cite|improve this question











$endgroup$



closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 6




    $begingroup$
    Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
    $endgroup$
    – RRL
    Jan 9 at 6:03






  • 1




    $begingroup$
    just try first few integers, and make a plot to see it. It will help you to get the idea
    $endgroup$
    – Lee
    Jan 9 at 6:15






  • 5




    $begingroup$
    See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:16






  • 1




    $begingroup$
    FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:53








  • 1




    $begingroup$
    You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
    $endgroup$
    – Zubzub
    Jan 9 at 7:48














0












0








0





$begingroup$


I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.



I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?










share|cite|improve this question











$endgroup$




I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.



I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?







calculus limits






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share|cite|improve this question













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share|cite|improve this question








edited Jan 9 at 6:27









Siong Thye Goh

104k1468120




104k1468120










asked Jan 9 at 5:56









Hami the PenguinHami the Penguin

474




474




closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 6




    $begingroup$
    Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
    $endgroup$
    – RRL
    Jan 9 at 6:03






  • 1




    $begingroup$
    just try first few integers, and make a plot to see it. It will help you to get the idea
    $endgroup$
    – Lee
    Jan 9 at 6:15






  • 5




    $begingroup$
    See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:16






  • 1




    $begingroup$
    FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:53








  • 1




    $begingroup$
    You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
    $endgroup$
    – Zubzub
    Jan 9 at 7:48














  • 6




    $begingroup$
    Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
    $endgroup$
    – RRL
    Jan 9 at 6:03






  • 1




    $begingroup$
    just try first few integers, and make a plot to see it. It will help you to get the idea
    $endgroup$
    – Lee
    Jan 9 at 6:15






  • 5




    $begingroup$
    See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:16






  • 1




    $begingroup$
    FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:53








  • 1




    $begingroup$
    You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
    $endgroup$
    – Zubzub
    Jan 9 at 7:48








6




6




$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03




$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03




1




1




$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15




$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15




5




5




$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16




$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16




1




1




$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53






$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53






1




1




$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48




$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48










3 Answers
3






active

oldest

votes


















4












$begingroup$

Hint: For the induction step prove and/or combine the following





  • $1+(n+1)<e(1+n)$,

  • the natural logarithm is an increasing function,


  • $ln(ex)=1+ln x $ for all $x$.




I assume that you haven't covered derivatives yet, and really need/want to do this by induction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
    $endgroup$
    – Jyrki Lahtonen
    Jan 9 at 6:39












  • $begingroup$
    Oh, I see how it works. Thank you very much.
    $endgroup$
    – Hami the Penguin
    Jan 10 at 0:14



















1












$begingroup$

Hint: Equivaently, show that $$1+n<e^n $$
for all $nge 1$. (Actually, this holds for all real $nne0$)






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint



    Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment



    This inequality is true for any $xge0$. The illustration below shows why:



    enter image description here






    share|cite|improve this answer











    $endgroup$




















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Hint: For the induction step prove and/or combine the following





      • $1+(n+1)<e(1+n)$,

      • the natural logarithm is an increasing function,


      • $ln(ex)=1+ln x $ for all $x$.




      I assume that you haven't covered derivatives yet, and really need/want to do this by induction.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
        $endgroup$
        – Jyrki Lahtonen
        Jan 9 at 6:39












      • $begingroup$
        Oh, I see how it works. Thank you very much.
        $endgroup$
        – Hami the Penguin
        Jan 10 at 0:14
















      4












      $begingroup$

      Hint: For the induction step prove and/or combine the following





      • $1+(n+1)<e(1+n)$,

      • the natural logarithm is an increasing function,


      • $ln(ex)=1+ln x $ for all $x$.




      I assume that you haven't covered derivatives yet, and really need/want to do this by induction.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
        $endgroup$
        – Jyrki Lahtonen
        Jan 9 at 6:39












      • $begingroup$
        Oh, I see how it works. Thank you very much.
        $endgroup$
        – Hami the Penguin
        Jan 10 at 0:14














      4












      4








      4





      $begingroup$

      Hint: For the induction step prove and/or combine the following





      • $1+(n+1)<e(1+n)$,

      • the natural logarithm is an increasing function,


      • $ln(ex)=1+ln x $ for all $x$.




      I assume that you haven't covered derivatives yet, and really need/want to do this by induction.






      share|cite|improve this answer











      $endgroup$



      Hint: For the induction step prove and/or combine the following





      • $1+(n+1)<e(1+n)$,

      • the natural logarithm is an increasing function,


      • $ln(ex)=1+ln x $ for all $x$.




      I assume that you haven't covered derivatives yet, and really need/want to do this by induction.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      answered Jan 9 at 6:34


























      community wiki





      Jyrki Lahtonen













      • $begingroup$
        Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
        $endgroup$
        – Jyrki Lahtonen
        Jan 9 at 6:39












      • $begingroup$
        Oh, I see how it works. Thank you very much.
        $endgroup$
        – Hami the Penguin
        Jan 10 at 0:14


















      • $begingroup$
        Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
        $endgroup$
        – Jyrki Lahtonen
        Jan 9 at 6:39












      • $begingroup$
        Oh, I see how it works. Thank you very much.
        $endgroup$
        – Hami the Penguin
        Jan 10 at 0:14
















      $begingroup$
      Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
      $endgroup$
      – Jyrki Lahtonen
      Jan 9 at 6:39






      $begingroup$
      Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
      $endgroup$
      – Jyrki Lahtonen
      Jan 9 at 6:39














      $begingroup$
      Oh, I see how it works. Thank you very much.
      $endgroup$
      – Hami the Penguin
      Jan 10 at 0:14




      $begingroup$
      Oh, I see how it works. Thank you very much.
      $endgroup$
      – Hami the Penguin
      Jan 10 at 0:14











      1












      $begingroup$

      Hint: Equivaently, show that $$1+n<e^n $$
      for all $nge 1$. (Actually, this holds for all real $nne0$)






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint: Equivaently, show that $$1+n<e^n $$
        for all $nge 1$. (Actually, this holds for all real $nne0$)






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint: Equivaently, show that $$1+n<e^n $$
          for all $nge 1$. (Actually, this holds for all real $nne0$)






          share|cite|improve this answer









          $endgroup$



          Hint: Equivaently, show that $$1+n<e^n $$
          for all $nge 1$. (Actually, this holds for all real $nne0$)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 7:29









          Hagen von EitzenHagen von Eitzen

          283k23273508




          283k23273508























              0












              $begingroup$

              Hint



              Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment



              This inequality is true for any $xge0$. The illustration below shows why:



              enter image description here






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Hint



                Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment



                This inequality is true for any $xge0$. The illustration below shows why:



                enter image description here






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint



                  Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment



                  This inequality is true for any $xge0$. The illustration below shows why:



                  enter image description here






                  share|cite|improve this answer











                  $endgroup$



                  Hint



                  Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment



                  This inequality is true for any $xge0$. The illustration below shows why:



                  enter image description here







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 9 at 7:49

























                  answered Jan 9 at 6:27









                  Mostafa AyazMostafa Ayaz

                  18.1k31040




                  18.1k31040















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