prove that $u_n$/$u_{n +1}$ ≥ 1.











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I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.



Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.



And $u_n$ = (1 + 1/n)$^{n+1}$



What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.






real-analysis







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    up vote
    -1
    down vote

    favorite












    I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.



    Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.



    And $u_n$ = (1 + 1/n)$^{n+1}$



    What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.






    real-analysis







    share|cite|improve this question







    New contributor




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






















      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.



      Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.



      And $u_n$ = (1 + 1/n)$^{n+1}$



      What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.






      real-analysis







      share|cite|improve this question







      New contributor




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











      I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.



      Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.



      And $u_n$ = (1 + 1/n)$^{n+1}$



      What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.






      real-analysis




      real-analysis






      share|cite|improve this question







      New contributor




      Peter van de Berg 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 question







      New contributor




      Peter van de Berg 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 question




      share|cite|improve this question






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      Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked yesterday









      Peter van de Berg

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          Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$



          Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$






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            Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$



            Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$



              Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$



                Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$






                share|cite|improve this answer












                Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$



                Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$







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                answered yesterday









                Sorin Tirc

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