Inequality $n^{sqrt{n+2}}>(n+1)^{sqrt{n+1}}$ solution

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How to find the solution of inequality
$$n^{sqrt{n+2}}>(n+1)^{sqrt{n+1}}$$
for integers with basic analysis without derivatives.



I don't have any idea how to start with it.






inequality







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$endgroup$












  • $begingroup$
    This is not true for $n=1$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:38










  • $begingroup$
    @Bo5man I'm looking for the integer from what it's true
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:42












  • $begingroup$
    From the statement I could not interpret that. Thanks for clarifying.
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:47
















2












$begingroup$


How to find the solution of inequality
$$n^{sqrt{n+2}}>(n+1)^{sqrt{n+1}}$$
for integers with basic analysis without derivatives.



I don't have any idea how to start with it.






inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is not true for $n=1$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:38










  • $begingroup$
    @Bo5man I'm looking for the integer from what it's true
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:42












  • $begingroup$
    From the statement I could not interpret that. Thanks for clarifying.
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:47














2












2








2





$begingroup$


How to find the solution of inequality
$$n^{sqrt{n+2}}>(n+1)^{sqrt{n+1}}$$
for integers with basic analysis without derivatives.



I don't have any idea how to start with it.






inequality







share|cite|improve this question











$endgroup$




How to find the solution of inequality
$$n^{sqrt{n+2}}>(n+1)^{sqrt{n+1}}$$
for integers with basic analysis without derivatives.



I don't have any idea how to start with it.






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 21:53







avan1235

















asked Dec 18 '18 at 21:26









avan1235avan1235

3297




3297












  • $begingroup$
    This is not true for $n=1$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:38










  • $begingroup$
    @Bo5man I'm looking for the integer from what it's true
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:42












  • $begingroup$
    From the statement I could not interpret that. Thanks for clarifying.
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:47


















  • $begingroup$
    This is not true for $n=1$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:38










  • $begingroup$
    @Bo5man I'm looking for the integer from what it's true
    $endgroup$
    – avan1235
    Dec 18 '18 at 21:42












  • $begingroup$
    From the statement I could not interpret that. Thanks for clarifying.
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:47
















$begingroup$
This is not true for $n=1$
$endgroup$
– Bo5man
Dec 18 '18 at 21:38




$begingroup$
This is not true for $n=1$
$endgroup$
– Bo5man
Dec 18 '18 at 21:38












$begingroup$
@Bo5man I'm looking for the integer from what it's true
$endgroup$
– avan1235
Dec 18 '18 at 21:42






$begingroup$
@Bo5man I'm looking for the integer from what it's true
$endgroup$
– avan1235
Dec 18 '18 at 21:42














$begingroup$
From the statement I could not interpret that. Thanks for clarifying.
$endgroup$
– Bo5man
Dec 18 '18 at 21:47




$begingroup$
From the statement I could not interpret that. Thanks for clarifying.
$endgroup$
– Bo5man
Dec 18 '18 at 21:47










1 Answer
1






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oldest

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2












$begingroup$

You are looking for the smallest $n$ such that



$$frac{n^{sqrt{n+2}}}{(n+1)^{sqrt{n+1}}} > 1$$ Take logarithms and search for the zero of
$$g(n)=sqrt{n+2} log (n)-sqrt{n+1} log (n+1)$$ Assume that $n$ is large and use
$$sqrt{n+2}sim sqrt n+frac 1 { sqrt n}$$
$$sqrt{n+1}sim sqrt n+frac 1 {2 sqrt n}$$
$$log(n+1)=log(n)+logleft(1+frac 1 nright)sim log(n)+frac 1 n$$ Replace to get
$$g(n) sim -frac{1}{2n^{3/2}} (2 n-n log (n)+1)$$
The solution of $2 n-n log (n)+1=0$ is given in terms of Lambert function; so, the approximation is
$$n =frac{1}{Wleft(frac{1}{e^2}right)}$$ Since the argument is small, use $W(x)approx frac{x}{x+1}$ which makes $color{red}{n=1+e^2 approx 8.39}$ while the exact solution of $g(n)=0$ in the real domain would be $8.70$. Then $color{red}{n=9}$.



Let us check the values of $$f(n)=n^{sqrt{n+2}}-(n+1)^{sqrt{n+1}}$$
$$f(8)=8^{sqrt{10}}-729 approx -11.50 qquad f(9)=9^{sqrt{11}}-10^{sqrt{10}}approx 8.68$$






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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    You are looking for the smallest $n$ such that



    $$frac{n^{sqrt{n+2}}}{(n+1)^{sqrt{n+1}}} > 1$$ Take logarithms and search for the zero of
    $$g(n)=sqrt{n+2} log (n)-sqrt{n+1} log (n+1)$$ Assume that $n$ is large and use
    $$sqrt{n+2}sim sqrt n+frac 1 { sqrt n}$$
    $$sqrt{n+1}sim sqrt n+frac 1 {2 sqrt n}$$
    $$log(n+1)=log(n)+logleft(1+frac 1 nright)sim log(n)+frac 1 n$$ Replace to get
    $$g(n) sim -frac{1}{2n^{3/2}} (2 n-n log (n)+1)$$
    The solution of $2 n-n log (n)+1=0$ is given in terms of Lambert function; so, the approximation is
    $$n =frac{1}{Wleft(frac{1}{e^2}right)}$$ Since the argument is small, use $W(x)approx frac{x}{x+1}$ which makes $color{red}{n=1+e^2 approx 8.39}$ while the exact solution of $g(n)=0$ in the real domain would be $8.70$. Then $color{red}{n=9}$.



    Let us check the values of $$f(n)=n^{sqrt{n+2}}-(n+1)^{sqrt{n+1}}$$
    $$f(8)=8^{sqrt{10}}-729 approx -11.50 qquad f(9)=9^{sqrt{11}}-10^{sqrt{10}}approx 8.68$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You are looking for the smallest $n$ such that



      $$frac{n^{sqrt{n+2}}}{(n+1)^{sqrt{n+1}}} > 1$$ Take logarithms and search for the zero of
      $$g(n)=sqrt{n+2} log (n)-sqrt{n+1} log (n+1)$$ Assume that $n$ is large and use
      $$sqrt{n+2}sim sqrt n+frac 1 { sqrt n}$$
      $$sqrt{n+1}sim sqrt n+frac 1 {2 sqrt n}$$
      $$log(n+1)=log(n)+logleft(1+frac 1 nright)sim log(n)+frac 1 n$$ Replace to get
      $$g(n) sim -frac{1}{2n^{3/2}} (2 n-n log (n)+1)$$
      The solution of $2 n-n log (n)+1=0$ is given in terms of Lambert function; so, the approximation is
      $$n =frac{1}{Wleft(frac{1}{e^2}right)}$$ Since the argument is small, use $W(x)approx frac{x}{x+1}$ which makes $color{red}{n=1+e^2 approx 8.39}$ while the exact solution of $g(n)=0$ in the real domain would be $8.70$. Then $color{red}{n=9}$.



      Let us check the values of $$f(n)=n^{sqrt{n+2}}-(n+1)^{sqrt{n+1}}$$
      $$f(8)=8^{sqrt{10}}-729 approx -11.50 qquad f(9)=9^{sqrt{11}}-10^{sqrt{10}}approx 8.68$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You are looking for the smallest $n$ such that



        $$frac{n^{sqrt{n+2}}}{(n+1)^{sqrt{n+1}}} > 1$$ Take logarithms and search for the zero of
        $$g(n)=sqrt{n+2} log (n)-sqrt{n+1} log (n+1)$$ Assume that $n$ is large and use
        $$sqrt{n+2}sim sqrt n+frac 1 { sqrt n}$$
        $$sqrt{n+1}sim sqrt n+frac 1 {2 sqrt n}$$
        $$log(n+1)=log(n)+logleft(1+frac 1 nright)sim log(n)+frac 1 n$$ Replace to get
        $$g(n) sim -frac{1}{2n^{3/2}} (2 n-n log (n)+1)$$
        The solution of $2 n-n log (n)+1=0$ is given in terms of Lambert function; so, the approximation is
        $$n =frac{1}{Wleft(frac{1}{e^2}right)}$$ Since the argument is small, use $W(x)approx frac{x}{x+1}$ which makes $color{red}{n=1+e^2 approx 8.39}$ while the exact solution of $g(n)=0$ in the real domain would be $8.70$. Then $color{red}{n=9}$.



        Let us check the values of $$f(n)=n^{sqrt{n+2}}-(n+1)^{sqrt{n+1}}$$
        $$f(8)=8^{sqrt{10}}-729 approx -11.50 qquad f(9)=9^{sqrt{11}}-10^{sqrt{10}}approx 8.68$$






        share|cite|improve this answer











        $endgroup$



        You are looking for the smallest $n$ such that



        $$frac{n^{sqrt{n+2}}}{(n+1)^{sqrt{n+1}}} > 1$$ Take logarithms and search for the zero of
        $$g(n)=sqrt{n+2} log (n)-sqrt{n+1} log (n+1)$$ Assume that $n$ is large and use
        $$sqrt{n+2}sim sqrt n+frac 1 { sqrt n}$$
        $$sqrt{n+1}sim sqrt n+frac 1 {2 sqrt n}$$
        $$log(n+1)=log(n)+logleft(1+frac 1 nright)sim log(n)+frac 1 n$$ Replace to get
        $$g(n) sim -frac{1}{2n^{3/2}} (2 n-n log (n)+1)$$
        The solution of $2 n-n log (n)+1=0$ is given in terms of Lambert function; so, the approximation is
        $$n =frac{1}{Wleft(frac{1}{e^2}right)}$$ Since the argument is small, use $W(x)approx frac{x}{x+1}$ which makes $color{red}{n=1+e^2 approx 8.39}$ while the exact solution of $g(n)=0$ in the real domain would be $8.70$. Then $color{red}{n=9}$.



        Let us check the values of $$f(n)=n^{sqrt{n+2}}-(n+1)^{sqrt{n+1}}$$
        $$f(8)=8^{sqrt{10}}-729 approx -11.50 qquad f(9)=9^{sqrt{11}}-10^{sqrt{10}}approx 8.68$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 10:32

























        answered Dec 19 '18 at 10:18









        Claude LeiboviciClaude Leibovici

        121k1157134




        121k1157134






























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