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












0














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




















  • 3




    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    – Yanko
    2 days ago






  • 1




    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    – Song
    2 days ago






  • 1




    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    – Clayton
    2 days ago










  • Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    – Shashwat1337
    2 days ago








  • 2




    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 $left(frac{n+1}{2}right)^n geqslant n!$
    – Calum Gilhooley
    2 days ago
















0














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




















  • 3




    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    – Yanko
    2 days ago






  • 1




    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    – Song
    2 days ago






  • 1




    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    – Clayton
    2 days ago










  • Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    – Shashwat1337
    2 days ago








  • 2




    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 $left(frac{n+1}{2}right)^n geqslant n!$
    – Calum Gilhooley
    2 days ago














0












0








0







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















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 2 days ago







Shashwat1337

















asked 2 days ago









Shashwat1337Shashwat1337

215




215








  • 3




    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    – Yanko
    2 days ago






  • 1




    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    – Song
    2 days ago






  • 1




    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    – Clayton
    2 days ago










  • Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    – Shashwat1337
    2 days ago








  • 2




    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 $left(frac{n+1}{2}right)^n geqslant n!$
    – Calum Gilhooley
    2 days ago














  • 3




    That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
    – Yanko
    2 days ago






  • 1




    It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
    – Song
    2 days ago






  • 1




    Actually, I think your inequality should be reversed because of Stirling's approximation formula.
    – Clayton
    2 days ago










  • Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
    – Shashwat1337
    2 days ago








  • 2




    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 $left(frac{n+1}{2}right)^n geqslant n!$
    – Calum Gilhooley
    2 days ago








3




3




That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
– Yanko
2 days ago




That's wrong. For $x=6$ you have $5!=120<46,656=3^6$
– Yanko
2 days ago




1




1




It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
– Song
2 days ago




It is not true since $n! sim sqrt{2pi n}(n/e)^n$ by Stirling's formula.
– Song
2 days ago




1




1




Actually, I think your inequality should be reversed because of Stirling's approximation formula.
– Clayton
2 days ago




Actually, I think your inequality should be reversed because of Stirling's approximation formula.
– Clayton
2 days ago












Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
– Shashwat1337
2 days ago






Yes sorry the reversed inequality holds. Also it is $(x/2)^{x-1}$ instead of $(x/2)^x$.
– Shashwat1337
2 days ago






2




2




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 $left(frac{n+1}{2}right)^n geqslant n!$
– Calum Gilhooley
2 days ago




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 $left(frac{n+1}{2}right)^n geqslant n!$
– Calum Gilhooley
2 days ago










2 Answers
2






active

oldest

votes


















1














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





















  • Thank you very much. The proof was great!
    – Shashwat1337
    2 days ago



















1














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























  • Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    – Shashwat1337
    2 days ago










  • You're welcome. Wish you luck!
    – Mostafa Ayaz
    2 days ago











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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





















  • Thank you very much. The proof was great!
    – Shashwat1337
    2 days ago
















1














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





















  • Thank you very much. The proof was great!
    – Shashwat1337
    2 days ago














1












1








1






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












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 2 days ago









Calum GilhooleyCalum Gilhooley

4,157529




4,157529












  • Thank you very much. The proof was great!
    – Shashwat1337
    2 days ago


















  • Thank you very much. The proof was great!
    – Shashwat1337
    2 days ago
















Thank you very much. The proof was great!
– Shashwat1337
2 days ago




Thank you very much. The proof was great!
– Shashwat1337
2 days ago











1














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























  • Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    – Shashwat1337
    2 days ago










  • You're welcome. Wish you luck!
    – Mostafa Ayaz
    2 days ago
















1














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























  • Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    – Shashwat1337
    2 days ago










  • You're welcome. Wish you luck!
    – Mostafa Ayaz
    2 days ago














1












1








1






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














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 2 days ago

























answered 2 days ago









Mostafa AyazMostafa Ayaz

14.1k3937




14.1k3937












  • Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    – Shashwat1337
    2 days ago










  • You're welcome. Wish you luck!
    – Mostafa Ayaz
    2 days ago


















  • Yes, I made a mistake. The reversed inequality holds. Thank you very much.
    – Shashwat1337
    2 days ago










  • You're welcome. Wish you luck!
    – Mostafa Ayaz
    2 days ago
















Yes, I made a mistake. The reversed inequality holds. Thank you very much.
– Shashwat1337
2 days ago




Yes, I made a mistake. The reversed inequality holds. Thank you very much.
– Shashwat1337
2 days ago












You're welcome. Wish you luck!
– Mostafa Ayaz
2 days ago




You're welcome. Wish you luck!
– Mostafa Ayaz
2 days ago


















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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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