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







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  • 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

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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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Antonio Litta Visconti Arese

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Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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