Proving with induction $(1-x)^n<frac 1 {1+nx}$












2












$begingroup$



Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$




My attempt:



Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.



Suppose the statement is true for $n$, prove for $n+1$:



$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$



Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?



Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?






induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    The first $to$ should rather be an $iff$.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:52










  • $begingroup$
    @PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:53








  • 1




    $begingroup$
    You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:54












  • $begingroup$
    Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:57










  • $begingroup$
    Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:57


















2












$begingroup$



Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$




My attempt:



Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.



Suppose the statement is true for $n$, prove for $n+1$:



$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$



Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?



Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?






induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    The first $to$ should rather be an $iff$.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:52










  • $begingroup$
    @PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:53








  • 1




    $begingroup$
    You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:54












  • $begingroup$
    Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:57










  • $begingroup$
    Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:57
















2












2








2





$begingroup$



Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$




My attempt:



Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.



Suppose the statement is true for $n$, prove for $n+1$:



$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$



Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?



Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?






induction







share|cite|improve this question











$endgroup$





Prove using induction that $forall ninmathbb N, forall xin mathbb R: 0<x<1: (1-x)^n<frac 1 {1+nx}$




My attempt:



Base: for $n=1: 1-x<frac 1 {1+x}iff 1-x^2<1$, true since $0<x<1$.



Suppose the statement is true for $n$, prove for $n+1$:



$(1-x)^{n+1}=(1-x)(1-x)^{n}overset{i.h}<frac{(1-x)}{1+nx}$



Now I got stuck, maybe another induction to show that $1+nx+x<1+nx$? Is there another way?



Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?






induction




induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 '15 at 21:26







shinzou

















asked Jan 28 '15 at 20:51









shinzoushinzou

2,0561635




2,0561635












  • $begingroup$
    The first $to$ should rather be an $iff$.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:52










  • $begingroup$
    @PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:53








  • 1




    $begingroup$
    You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:54












  • $begingroup$
    Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:57










  • $begingroup$
    Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:57




















  • $begingroup$
    The first $to$ should rather be an $iff$.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:52










  • $begingroup$
    @PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:53








  • 1




    $begingroup$
    You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:54












  • $begingroup$
    Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
    $endgroup$
    – shinzou
    Jan 28 '15 at 20:57










  • $begingroup$
    Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 28 '15 at 20:57


















$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52




$begingroup$
The first $to$ should rather be an $iff$.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:52












$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53






$begingroup$
@PedroTamaroff why is it wrong to write just $rightarrow$? isn't it enough for this case?
$endgroup$
– shinzou
Jan 28 '15 at 20:53






1




1




$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54






$begingroup$
You want to show that if $0<x<1$, $1-x<frac{1}{1+x}$. You cannot assume what you want to show is true, show it implies a true fact and conclude that what you assumed is true is... true. Your logic is "Let's show $P$. Assume $P$. Since $P$ implies $Q$ and $Q$ is true, $P$ is true."
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:54














$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57




$begingroup$
Oh yes I see that $frac{1-x}{1+nx}<frac{1}{1+nx}$ the problem is with the denominator.
$endgroup$
– shinzou
Jan 28 '15 at 20:57












$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57






$begingroup$
Note that $$frac{1-x}{1+nx}<frac{1}{1+(n+1)x}$$ is true if and only if $-(n+1)x^2<0$ (just cross multiply and play around a while). Since $x^2>0$ and $n+1>0$, you're done.
$endgroup$
– Pedro Tamaroff♦
Jan 28 '15 at 20:57












3 Answers
3






active

oldest

votes


















5












$begingroup$

Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get



$$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Is there nothing wrong with starting with $(1-x)^{n+1}$?
    $endgroup$
    – shinzou
    Jan 28 '15 at 21:00





















2












$begingroup$

Notice that there is no need of induction. By the AM-GM inequality,
$$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
    $endgroup$
    – marty cohen
    Jan 10 at 21:59












  • $begingroup$
    The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
    $endgroup$
    – marty cohen
    Jan 10 at 22:03



















0












$begingroup$

Write it in the form
$(1+nx)(1-x)^n
lt 1$.



If true for $n$, then



$begin{array}\
(1+(n+1)x)(1-x)^{n+1}
&=(1+(n+1)x)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
&=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
&<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
&=dfrac{1+nx-(n+1)x^2}{1+nx}\
&lt 1\
end{array}
$






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get



    $$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Is there nothing wrong with starting with $(1-x)^{n+1}$?
      $endgroup$
      – shinzou
      Jan 28 '15 at 21:00


















    5












    $begingroup$

    Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get



    $$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Is there nothing wrong with starting with $(1-x)^{n+1}$?
      $endgroup$
      – shinzou
      Jan 28 '15 at 21:00
















    5












    5








    5





    $begingroup$

    Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get



    $$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$






    share|cite|improve this answer









    $endgroup$



    Apply again the base case: $1-x<displaystylefrac1{1+x}$ and that $x^2>0$ to get



    $$frac{1-x}{1+nx} < frac1{(1+nx)(1+x)}=frac1{1+(n+1)x+nx^2}<frac1{1+(n+1)x},.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 28 '15 at 20:57









    BerciBerci

    61.3k23674




    61.3k23674












    • $begingroup$
      Is there nothing wrong with starting with $(1-x)^{n+1}$?
      $endgroup$
      – shinzou
      Jan 28 '15 at 21:00




















    • $begingroup$
      Is there nothing wrong with starting with $(1-x)^{n+1}$?
      $endgroup$
      – shinzou
      Jan 28 '15 at 21:00


















    $begingroup$
    Is there nothing wrong with starting with $(1-x)^{n+1}$?
    $endgroup$
    – shinzou
    Jan 28 '15 at 21:00






    $begingroup$
    Is there nothing wrong with starting with $(1-x)^{n+1}$?
    $endgroup$
    – shinzou
    Jan 28 '15 at 21:00













    2












    $begingroup$

    Notice that there is no need of induction. By the AM-GM inequality,
    $$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
      $endgroup$
      – marty cohen
      Jan 10 at 21:59












    • $begingroup$
      The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
      $endgroup$
      – marty cohen
      Jan 10 at 22:03
















    2












    $begingroup$

    Notice that there is no need of induction. By the AM-GM inequality,
    $$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
      $endgroup$
      – marty cohen
      Jan 10 at 21:59












    • $begingroup$
      The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
      $endgroup$
      – marty cohen
      Jan 10 at 22:03














    2












    2








    2





    $begingroup$

    Notice that there is no need of induction. By the AM-GM inequality,
    $$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$






    share|cite|improve this answer









    $endgroup$



    Notice that there is no need of induction. By the AM-GM inequality,
    $$ (1+nx)(1-x)^n < left(frac{(1+nx)+n(1-x)}{n+1}right)^{n+1} =1.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 28 '15 at 22:15









    Jack D'AurizioJack D'Aurizio

    290k33284666




    290k33284666








    • 1




      $begingroup$
      Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
      $endgroup$
      – marty cohen
      Jan 10 at 21:59












    • $begingroup$
      The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
      $endgroup$
      – marty cohen
      Jan 10 at 22:03














    • 1




      $begingroup$
      Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
      $endgroup$
      – marty cohen
      Jan 10 at 21:59












    • $begingroup$
      The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
      $endgroup$
      – marty cohen
      Jan 10 at 22:03








    1




    1




    $begingroup$
    Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
    $endgroup$
    – marty cohen
    Jan 10 at 21:59






    $begingroup$
    Ingenious. Similar to the proofs that $(1+1/n)^n$ is increasing and $(1+1/n)^{n+1}$ is decreasing (and so both approach a common limit, which I propose to call "$e$").
    $endgroup$
    – marty cohen
    Jan 10 at 21:59














    $begingroup$
    The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
    $endgroup$
    – marty cohen
    Jan 10 at 22:03




    $begingroup$
    The proofs are in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.
    $endgroup$
    – marty cohen
    Jan 10 at 22:03











    0












    $begingroup$

    Write it in the form
    $(1+nx)(1-x)^n
    lt 1$    .



    If true for $n$, then



    $begin{array}\
    (1+(n+1)x)(1-x)^{n+1}
    &=(1+(n+1)x)(1-x)^{n+1}\
    &=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
    &=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
    &<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
    &=dfrac{1+nx-(n+1)x^2}{1+nx}\
    &lt 1\
    end{array}
    $






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Write it in the form
      $(1+nx)(1-x)^n
      lt 1$      .



      If true for $n$, then



      $begin{array}\
      (1+(n+1)x)(1-x)^{n+1}
      &=(1+(n+1)x)(1-x)^{n+1}\
      &=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
      &=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
      &<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
      &=dfrac{1+nx-(n+1)x^2}{1+nx}\
      &lt 1\
      end{array}
      $






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Write it in the form
        $(1+nx)(1-x)^n
        lt 1$        .



        If true for $n$, then



        $begin{array}\
        (1+(n+1)x)(1-x)^{n+1}
        &=(1+(n+1)x)(1-x)^{n+1}\
        &=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
        &=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
        &<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
        &=dfrac{1+nx-(n+1)x^2}{1+nx}\
        &lt 1\
        end{array}
        $






        share|cite|improve this answer









        $endgroup$



        Write it in the form
        $(1+nx)(1-x)^n
        lt 1$        .



        If true for $n$, then



        $begin{array}\
        (1+(n+1)x)(1-x)^{n+1}
        &=(1+(n+1)x)(1-x)^{n+1}\
        &=dfrac{(1+(n+1)x)}{1+nx}(1+nx)(1-x)^{n+1}\
        &=dfrac{(1+(n+1)x)(1-x)}{1+nx}(1+nx)(1-x)^{n}\
        &<dfrac{(1+(n+1)x)(1-x)}{1+nx}\
        &=dfrac{1+nx-(n+1)x^2}{1+nx}\
        &lt 1\
        end{array}
        $







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Jan 10 at 22:08









        marty cohenmarty cohen

        74.2k549128




        74.2k549128






























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