I have proved that $frac{n!}{n^n}leqfrac{2}{n^2}$ But I don't know how one came up with it












0












$begingroup$


I have proved by induction that the Statement above is true



Here is my proof:



Inductionbase: $frac{2!}{2^2}=frac{2}{4}$



Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$



But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:



$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$






real-analysis proof-explanation







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
    $endgroup$
    – User8128
    Jan 30 at 17:20






  • 2




    $begingroup$
    In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
    $endgroup$
    – New2Math
    Jan 30 at 17:23
















0












$begingroup$


I have proved by induction that the Statement above is true



Here is my proof:



Inductionbase: $frac{2!}{2^2}=frac{2}{4}$



Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$



But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:



$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$






real-analysis proof-explanation







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
    $endgroup$
    – User8128
    Jan 30 at 17:20






  • 2




    $begingroup$
    In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
    $endgroup$
    – New2Math
    Jan 30 at 17:23














0












0








0


1



$begingroup$


I have proved by induction that the Statement above is true



Here is my proof:



Inductionbase: $frac{2!}{2^2}=frac{2}{4}$



Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$



But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:



$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$






real-analysis proof-explanation







share|cite|improve this question









$endgroup$




I have proved by induction that the Statement above is true



Here is my proof:



Inductionbase: $frac{2!}{2^2}=frac{2}{4}$



Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$



But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:



$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$






real-analysis proof-explanation




real-analysis proof-explanation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 30 at 17:14









New2MathNew2Math

16715




16715








  • 3




    $begingroup$
    This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
    $endgroup$
    – User8128
    Jan 30 at 17:20






  • 2




    $begingroup$
    In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
    $endgroup$
    – New2Math
    Jan 30 at 17:23














  • 3




    $begingroup$
    This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
    $endgroup$
    – User8128
    Jan 30 at 17:20






  • 2




    $begingroup$
    In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
    $endgroup$
    – New2Math
    Jan 30 at 17:23








3




3




$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20




$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20




2




2




$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23




$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23










2 Answers
2






active

oldest

votes


















4












$begingroup$

I would write the inequality in the form
$$n!le 2cdot n^{n-2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
    $endgroup$
    – New2Math
    Jan 30 at 17:33



















1












$begingroup$

Your proof is invalid, because you have to show that




if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$




Now
$$
frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
$$
by the induction hypothesis.



You can finish by proving that, for $nge2$,
$$
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
$$
that's equivalent to
$$
frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
$$
or, as well, to
$$
left(frac{n}{n+1}right)^{n-2}le1
$$
which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).



Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
$$
frac{1!}{1^1}lefrac{2}{1^2}
qquad
frac{2!}{2^2}lefrac{2}{2^2}
$$
you're done.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    I would write the inequality in the form
    $$n!le 2cdot n^{n-2}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
      $endgroup$
      – New2Math
      Jan 30 at 17:33
















    4












    $begingroup$

    I would write the inequality in the form
    $$n!le 2cdot n^{n-2}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
      $endgroup$
      – New2Math
      Jan 30 at 17:33














    4












    4








    4





    $begingroup$

    I would write the inequality in the form
    $$n!le 2cdot n^{n-2}$$






    share|cite|improve this answer









    $endgroup$



    I would write the inequality in the form
    $$n!le 2cdot n^{n-2}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 at 17:17









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    78.4k42867




    78.4k42867












    • $begingroup$
      I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
      $endgroup$
      – New2Math
      Jan 30 at 17:33


















    • $begingroup$
      I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
      $endgroup$
      – New2Math
      Jan 30 at 17:33
















    $begingroup$
    I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
    $endgroup$
    – New2Math
    Jan 30 at 17:33




    $begingroup$
    I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
    $endgroup$
    – New2Math
    Jan 30 at 17:33











    1












    $begingroup$

    Your proof is invalid, because you have to show that




    if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$




    Now
    $$
    frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
    dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
    $$
    by the induction hypothesis.



    You can finish by proving that, for $nge2$,
    $$
    dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
    $$
    that's equivalent to
    $$
    frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
    $$
    or, as well, to
    $$
    left(frac{n}{n+1}right)^{n-2}le1
    $$
    which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).



    Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
    $$
    frac{1!}{1^1}lefrac{2}{1^2}
    qquad
    frac{2!}{2^2}lefrac{2}{2^2}
    $$
    you're done.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Your proof is invalid, because you have to show that




      if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$




      Now
      $$
      frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
      dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
      $$
      by the induction hypothesis.



      You can finish by proving that, for $nge2$,
      $$
      dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
      $$
      that's equivalent to
      $$
      frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
      $$
      or, as well, to
      $$
      left(frac{n}{n+1}right)^{n-2}le1
      $$
      which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).



      Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
      $$
      frac{1!}{1^1}lefrac{2}{1^2}
      qquad
      frac{2!}{2^2}lefrac{2}{2^2}
      $$
      you're done.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Your proof is invalid, because you have to show that




        if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$




        Now
        $$
        frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
        dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
        $$
        by the induction hypothesis.



        You can finish by proving that, for $nge2$,
        $$
        dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
        $$
        that's equivalent to
        $$
        frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
        $$
        or, as well, to
        $$
        left(frac{n}{n+1}right)^{n-2}le1
        $$
        which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).



        Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
        $$
        frac{1!}{1^1}lefrac{2}{1^2}
        qquad
        frac{2!}{2^2}lefrac{2}{2^2}
        $$
        you're done.






        share|cite|improve this answer









        $endgroup$



        Your proof is invalid, because you have to show that




        if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$




        Now
        $$
        frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
        dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
        $$
        by the induction hypothesis.



        You can finish by proving that, for $nge2$,
        $$
        dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
        $$
        that's equivalent to
        $$
        frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
        $$
        or, as well, to
        $$
        left(frac{n}{n+1}right)^{n-2}le1
        $$
        which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).



        Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
        $$
        frac{1!}{1^1}lefrac{2}{1^2}
        qquad
        frac{2!}{2^2}lefrac{2}{2^2}
        $$
        you're done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 17:44









        egregegreg

        185k1486207




        185k1486207






























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