How do we prove that $(x-1)!leq{(frac{x}{2})^{x-1}}$?












0












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I plotted the graph of $y=(x-1)!$ and $y=(frac{x}{2})^{x-1}$ and found that the latter is always greater than the former for $x>2$. Also, equality holds at $x=2$. Would someone please help me to prove it mathematically?






inequality







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  • 3




    $begingroup$
    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    $endgroup$
    – Yanko
    Jan 5 at 18:29






  • 1




    $begingroup$
    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    $endgroup$
    – Song
    Jan 5 at 18:31






  • 1




    $begingroup$
    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    $endgroup$
    – Clayton
    Jan 5 at 18:32










  • $begingroup$
    Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:34








  • 2




    $begingroup$
    For every positive integer $n$, the arithmetic mean of $1, 2, ldots, n$ is $frac{n+1}{2}$, and the geometric mean of the same numbers is $sqrt[n]{n!}$, so $left(frac{n+1}{2}right)^n geqslant n!$
    $endgroup$
    – Calum Gilhooley
    Jan 5 at 18:39
















0












$begingroup$


I plotted the graph of $y=(x-1)!$ and $y=(frac{x}{2})^{x-1}$ and found that the latter is always greater than the former for $x>2$. Also, equality holds at $x=2$. Would someone please help me to prove it mathematically?






inequality







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    $endgroup$
    – Yanko
    Jan 5 at 18:29






  • 1




    $begingroup$
    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    $endgroup$
    – Song
    Jan 5 at 18:31






  • 1




    $begingroup$
    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    $endgroup$
    – Clayton
    Jan 5 at 18:32










  • $begingroup$
    Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:34








  • 2




    $begingroup$
    For every positive integer $n$, the arithmetic mean of $1, 2, ldots, n$ is $frac{n+1}{2}$, and the geometric mean of the same numbers is $sqrt[n]{n!}$, so $left(frac{n+1}{2}right)^n geqslant n!$
    $endgroup$
    – Calum Gilhooley
    Jan 5 at 18:39














0












0








0





$begingroup$


I plotted the graph of $y=(x-1)!$ and $y=(frac{x}{2})^{x-1}$ and found that the latter is always greater than the former for $x>2$. Also, equality holds at $x=2$. Would someone please help me to prove it mathematically?






inequality







share|cite|improve this question











$endgroup$




I plotted the graph of $y=(x-1)!$ and $y=(frac{x}{2})^{x-1}$ and found that the latter is always greater than the former for $x>2$. Also, equality holds at $x=2$. Would someone please help me to prove it mathematically?






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 9:37







Shashwat1337

















asked Jan 5 at 18:26









Shashwat1337Shashwat1337

959




959








  • 3




    $begingroup$
    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    $endgroup$
    – Yanko
    Jan 5 at 18:29






  • 1




    $begingroup$
    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    $endgroup$
    – Song
    Jan 5 at 18:31






  • 1




    $begingroup$
    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    $endgroup$
    – Clayton
    Jan 5 at 18:32










  • $begingroup$
    Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:34








  • 2




    $begingroup$
    For every positive integer $n$, the arithmetic mean of $1, 2, ldots, n$ is $frac{n+1}{2}$, and the geometric mean of the same numbers is $sqrt[n]{n!}$, so $left(frac{n+1}{2}right)^n geqslant n!$
    $endgroup$
    – Calum Gilhooley
    Jan 5 at 18:39














  • 3




    $begingroup$
    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    $endgroup$
    – Yanko
    Jan 5 at 18:29






  • 1




    $begingroup$
    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    $endgroup$
    – Song
    Jan 5 at 18:31






  • 1




    $begingroup$
    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    $endgroup$
    – Clayton
    Jan 5 at 18:32










  • $begingroup$
    Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:34








  • 2




    $begingroup$
    For every positive integer $n$, the arithmetic mean of $1, 2, ldots, n$ is $frac{n+1}{2}$, and the geometric mean of the same numbers is $sqrt[n]{n!}$, so $left(frac{n+1}{2}right)^n geqslant n!$
    $endgroup$
    – Calum Gilhooley
    Jan 5 at 18:39








3




3




$begingroup$
That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
$endgroup$
– Yanko
Jan 5 at 18:29




$begingroup$
That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
$endgroup$
– Yanko
Jan 5 at 18:29




1




1




$begingroup$
It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
$endgroup$
– Song
Jan 5 at 18:31




$begingroup$
It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
$endgroup$
– Song
Jan 5 at 18:31




1




1




$begingroup$
Actually, I think your inequality should be reversed because of Stirling's approximation formula.
$endgroup$
– Clayton
Jan 5 at 18:32




$begingroup$
Actually, I think your inequality should be reversed because of Stirling's approximation formula.
$endgroup$
– Clayton
Jan 5 at 18:32












$begingroup$
Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
$endgroup$
– Shashwat1337
Jan 5 at 18:34






$begingroup$
Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
$endgroup$
– Shashwat1337
Jan 5 at 18:34






2




2




$begingroup$
For every positive integer $n$, the arithmetic mean of $1, 2, ldots, n$ is $frac{n+1}{2}$, and the geometric mean of the same numbers is $sqrt[n]{n!}$, so $left(frac{n+1}{2}right)^n geqslant n!$
$endgroup$
– Calum Gilhooley
Jan 5 at 18:39




$begingroup$
For every positive integer $n$, the arithmetic mean of $1, 2, ldots, n$ is $frac{n+1}{2}$, and the geometric mean of the same numbers is $sqrt[n]{n!}$, so $left(frac{n+1}{2}right)^n geqslant n!$
$endgroup$
– Calum Gilhooley
Jan 5 at 18:39










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your conclusion is wrong. For $x=4$ we have $$(x-1)!=6\({xover 2})^x=16$$and $$(x-1)!notge ({xover 2})^x$$



Comment



The inequality is true in reverse side for large enough $n$ using Stirling's Approximation for factorial as follows $$(x-1)!<x!approx sqrt {2pi x}({xover e})^x<({xover 2})^x$$since $$sqrt{2pi x}<({eover 2})^x$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:42










  • $begingroup$
    You're welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 18:43



















1












$begingroup$

As already noted in a comment, if $x$ is intended only to have integral values $geqslant 2$, then the inequality (corrected now, so that the inequality sign points in the right direction) follows from the AM-GM inequality, applied to the $x - 1$ numbers $1, 2, ldots, x - 1$.



But the question is naturally read as applying to all real values of $x geqslant 2$, so for the sake of other readers it is also worth proving the inequality in the more general case, with $(x - 1)!$ understood to mean $Gamma(x)$.



To prove
begin{equation}
label{ineq:1}tag{1}
logGamma(x) leqslant (x - 1)logleft(frac{x}{2}right) quad (x geqslant 2),
end{equation}
because this holds for $x = 2$, it is enough to prove the inequality obtained by differentiating both sides:
begin{equation}
label{ineq:2}tag{2}
psi(x) leqslant logleft(frac{x}{2}right) + 1 - frac{1}{x} quad (x geqslant 2),
end{equation}
where $psi$ is the Digamma function.



As a special case of the formula
$$
psi(n) = H_{n-1} - gamma,
$$
we have
$$
psi(2) = 1 - gamma < frac{1}{2},
$$
so eqref{ineq:2} holds for $x = 2$.



Differentiating again, we find that it is enough to prove
begin{equation}
label{ineq:3}tag{3}
sum_{n=0}^inftyfrac{1}{(x + n)^2} leqslant frac{1}{x} + frac{1}{x^2} quad (x geqslant 2).
end{equation}
Indeed,
$$
sum_{n=1}^inftyfrac{1}{(x + n)^2} < sum_{n=1}^inftyfrac{1}{(x + n - 1)(x + n)} = sum_{n=1}^inftyleft(frac{1}{x + n - 1} - frac{1}{x + n}right) = frac{1}{x}.
$$
This proves eqref{ineq:3}, therefore eqref{ineq:2}, and therefore eqref{ineq:1}, with strict inequality for $x > 2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much. The proof was great!
    $endgroup$
    – Shashwat1337
    Jan 6 at 9:41











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your conclusion is wrong. For $x=4$ we have $$(x-1)!=6\({xover 2})^x=16$$and $$(x-1)!notge ({xover 2})^x$$



Comment



The inequality is true in reverse side for large enough $n$ using Stirling's Approximation for factorial as follows $$(x-1)!<x!approx sqrt {2pi x}({xover e})^x<({xover 2})^x$$since $$sqrt{2pi x}<({eover 2})^x$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:42










  • $begingroup$
    You're welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 18:43
















1












$begingroup$

Your conclusion is wrong. For $x=4$ we have $$(x-1)!=6\({xover 2})^x=16$$and $$(x-1)!notge ({xover 2})^x$$



Comment



The inequality is true in reverse side for large enough $n$ using Stirling's Approximation for factorial as follows $$(x-1)!<x!approx sqrt {2pi x}({xover e})^x<({xover 2})^x$$since $$sqrt{2pi x}<({eover 2})^x$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:42










  • $begingroup$
    You're welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 18:43














1












1








1





$begingroup$

Your conclusion is wrong. For $x=4$ we have $$(x-1)!=6\({xover 2})^x=16$$and $$(x-1)!notge ({xover 2})^x$$



Comment



The inequality is true in reverse side for large enough $n$ using Stirling's Approximation for factorial as follows $$(x-1)!<x!approx sqrt {2pi x}({xover e})^x<({xover 2})^x$$since $$sqrt{2pi x}<({eover 2})^x$$






share|cite|improve this answer











$endgroup$



Your conclusion is wrong. For $x=4$ we have $$(x-1)!=6\({xover 2})^x=16$$and $$(x-1)!notge ({xover 2})^x$$



Comment



The inequality is true in reverse side for large enough $n$ using Stirling's Approximation for factorial as follows $$(x-1)!<x!approx sqrt {2pi x}({xover e})^x<({xover 2})^x$$since $$sqrt{2pi x}<({eover 2})^x$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 18:44

























answered Jan 5 at 18:33









Mostafa AyazMostafa Ayaz

16.7k3939




16.7k3939












  • $begingroup$
    Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:42










  • $begingroup$
    You're welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 18:43


















  • $begingroup$
    Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    $endgroup$
    – Shashwat1337
    Jan 5 at 18:42










  • $begingroup$
    You're welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 18:43
















$begingroup$
Yes, I made a mistake. The reversed inequality holds. Thank you very much.
$endgroup$
– Shashwat1337
Jan 5 at 18:42




$begingroup$
Yes, I made a mistake. The reversed inequality holds. Thank you very much.
$endgroup$
– Shashwat1337
Jan 5 at 18:42












$begingroup$
You're welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 5 at 18:43




$begingroup$
You're welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 5 at 18:43











1












$begingroup$

As already noted in a comment, if $x$ is intended only to have integral values $geqslant 2$, then the inequality (corrected now, so that the inequality sign points in the right direction) follows from the AM-GM inequality, applied to the $x - 1$ numbers $1, 2, ldots, x - 1$.



But the question is naturally read as applying to all real values of $x geqslant 2$, so for the sake of other readers it is also worth proving the inequality in the more general case, with $(x - 1)!$ understood to mean $Gamma(x)$.



To prove
begin{equation}
label{ineq:1}tag{1}
logGamma(x) leqslant (x - 1)logleft(frac{x}{2}right) quad (x geqslant 2),
end{equation}
because this holds for $x = 2$, it is enough to prove the inequality obtained by differentiating both sides:
begin{equation}
label{ineq:2}tag{2}
psi(x) leqslant logleft(frac{x}{2}right) + 1 - frac{1}{x} quad (x geqslant 2),
end{equation}
where $psi$ is the Digamma function.



As a special case of the formula
$$
psi(n) = H_{n-1} - gamma,
$$
we have
$$
psi(2) = 1 - gamma < frac{1}{2},
$$
so eqref{ineq:2} holds for $x = 2$.



Differentiating again, we find that it is enough to prove
begin{equation}
label{ineq:3}tag{3}
sum_{n=0}^inftyfrac{1}{(x + n)^2} leqslant frac{1}{x} + frac{1}{x^2} quad (x geqslant 2).
end{equation}
Indeed,
$$
sum_{n=1}^inftyfrac{1}{(x + n)^2} < sum_{n=1}^inftyfrac{1}{(x + n - 1)(x + n)} = sum_{n=1}^inftyleft(frac{1}{x + n - 1} - frac{1}{x + n}right) = frac{1}{x}.
$$
This proves eqref{ineq:3}, therefore eqref{ineq:2}, and therefore eqref{ineq:1}, with strict inequality for $x > 2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much. The proof was great!
    $endgroup$
    – Shashwat1337
    Jan 6 at 9:41
















1












$begingroup$

As already noted in a comment, if $x$ is intended only to have integral values $geqslant 2$, then the inequality (corrected now, so that the inequality sign points in the right direction) follows from the AM-GM inequality, applied to the $x - 1$ numbers $1, 2, ldots, x - 1$.



But the question is naturally read as applying to all real values of $x geqslant 2$, so for the sake of other readers it is also worth proving the inequality in the more general case, with $(x - 1)!$ understood to mean $Gamma(x)$.



To prove
begin{equation}
label{ineq:1}tag{1}
logGamma(x) leqslant (x - 1)logleft(frac{x}{2}right) quad (x geqslant 2),
end{equation}
because this holds for $x = 2$, it is enough to prove the inequality obtained by differentiating both sides:
begin{equation}
label{ineq:2}tag{2}
psi(x) leqslant logleft(frac{x}{2}right) + 1 - frac{1}{x} quad (x geqslant 2),
end{equation}
where $psi$ is the Digamma function.



As a special case of the formula
$$
psi(n) = H_{n-1} - gamma,
$$
we have
$$
psi(2) = 1 - gamma < frac{1}{2},
$$
so eqref{ineq:2} holds for $x = 2$.



Differentiating again, we find that it is enough to prove
begin{equation}
label{ineq:3}tag{3}
sum_{n=0}^inftyfrac{1}{(x + n)^2} leqslant frac{1}{x} + frac{1}{x^2} quad (x geqslant 2).
end{equation}
Indeed,
$$
sum_{n=1}^inftyfrac{1}{(x + n)^2} < sum_{n=1}^inftyfrac{1}{(x + n - 1)(x + n)} = sum_{n=1}^inftyleft(frac{1}{x + n - 1} - frac{1}{x + n}right) = frac{1}{x}.
$$
This proves eqref{ineq:3}, therefore eqref{ineq:2}, and therefore eqref{ineq:1}, with strict inequality for $x > 2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much. The proof was great!
    $endgroup$
    – Shashwat1337
    Jan 6 at 9:41














1












1








1





$begingroup$

As already noted in a comment, if $x$ is intended only to have integral values $geqslant 2$, then the inequality (corrected now, so that the inequality sign points in the right direction) follows from the AM-GM inequality, applied to the $x - 1$ numbers $1, 2, ldots, x - 1$.



But the question is naturally read as applying to all real values of $x geqslant 2$, so for the sake of other readers it is also worth proving the inequality in the more general case, with $(x - 1)!$ understood to mean $Gamma(x)$.



To prove
begin{equation}
label{ineq:1}tag{1}
logGamma(x) leqslant (x - 1)logleft(frac{x}{2}right) quad (x geqslant 2),
end{equation}
because this holds for $x = 2$, it is enough to prove the inequality obtained by differentiating both sides:
begin{equation}
label{ineq:2}tag{2}
psi(x) leqslant logleft(frac{x}{2}right) + 1 - frac{1}{x} quad (x geqslant 2),
end{equation}
where $psi$ is the Digamma function.



As a special case of the formula
$$
psi(n) = H_{n-1} - gamma,
$$
we have
$$
psi(2) = 1 - gamma < frac{1}{2},
$$
so eqref{ineq:2} holds for $x = 2$.



Differentiating again, we find that it is enough to prove
begin{equation}
label{ineq:3}tag{3}
sum_{n=0}^inftyfrac{1}{(x + n)^2} leqslant frac{1}{x} + frac{1}{x^2} quad (x geqslant 2).
end{equation}
Indeed,
$$
sum_{n=1}^inftyfrac{1}{(x + n)^2} < sum_{n=1}^inftyfrac{1}{(x + n - 1)(x + n)} = sum_{n=1}^inftyleft(frac{1}{x + n - 1} - frac{1}{x + n}right) = frac{1}{x}.
$$
This proves eqref{ineq:3}, therefore eqref{ineq:2}, and therefore eqref{ineq:1}, with strict inequality for $x > 2$.






share|cite|improve this answer









$endgroup$



As already noted in a comment, if $x$ is intended only to have integral values $geqslant 2$, then the inequality (corrected now, so that the inequality sign points in the right direction) follows from the AM-GM inequality, applied to the $x - 1$ numbers $1, 2, ldots, x - 1$.



But the question is naturally read as applying to all real values of $x geqslant 2$, so for the sake of other readers it is also worth proving the inequality in the more general case, with $(x - 1)!$ understood to mean $Gamma(x)$.



To prove
begin{equation}
label{ineq:1}tag{1}
logGamma(x) leqslant (x - 1)logleft(frac{x}{2}right) quad (x geqslant 2),
end{equation}
because this holds for $x = 2$, it is enough to prove the inequality obtained by differentiating both sides:
begin{equation}
label{ineq:2}tag{2}
psi(x) leqslant logleft(frac{x}{2}right) + 1 - frac{1}{x} quad (x geqslant 2),
end{equation}
where $psi$ is the Digamma function.



As a special case of the formula
$$
psi(n) = H_{n-1} - gamma,
$$
we have
$$
psi(2) = 1 - gamma < frac{1}{2},
$$
so eqref{ineq:2} holds for $x = 2$.



Differentiating again, we find that it is enough to prove
begin{equation}
label{ineq:3}tag{3}
sum_{n=0}^inftyfrac{1}{(x + n)^2} leqslant frac{1}{x} + frac{1}{x^2} quad (x geqslant 2).
end{equation}
Indeed,
$$
sum_{n=1}^inftyfrac{1}{(x + n)^2} < sum_{n=1}^inftyfrac{1}{(x + n - 1)(x + n)} = sum_{n=1}^inftyleft(frac{1}{x + n - 1} - frac{1}{x + n}right) = frac{1}{x}.
$$
This proves eqref{ineq:3}, therefore eqref{ineq:2}, and therefore eqref{ineq:1}, with strict inequality for $x > 2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 21:38









Calum GilhooleyCalum Gilhooley

4,929630




4,929630












  • $begingroup$
    Thank you very much. The proof was great!
    $endgroup$
    – Shashwat1337
    Jan 6 at 9:41


















  • $begingroup$
    Thank you very much. The proof was great!
    $endgroup$
    – Shashwat1337
    Jan 6 at 9:41
















$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Jan 6 at 9:41




$begingroup$
Thank you very much. The proof was great!
$endgroup$
– Shashwat1337
Jan 6 at 9:41


















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