Proof of a factorial inequality [closed]
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How can I prove the following inequality?
$$sqrt{n+dfrac{1}{4}}ledfrac{n!}{left(n-dfrac{1}{2}right)!}lesqrt{n+dfrac{1}{4}+dfrac{1}{16n+12}}$$
Thanks
inequality
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
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closed as off-topic by A. Pongrácz, Chris Custer, José Carlos Santos, Cesareo, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can I prove the following inequality?
$$sqrt{n+dfrac{1}{4}}ledfrac{n!}{left(n-dfrac{1}{2}right)!}lesqrt{n+dfrac{1}{4}+dfrac{1}{16n+12}}$$
Thanks
inequality
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
$endgroup$
closed as off-topic by A. Pongrácz, Chris Custer, José Carlos Santos, Cesareo, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – A. Pongrácz, Chris Custer, José Carlos Santos, Cesareo, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
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$n in mathbb{N}$? What is $(n-1/2)!$ then?
$endgroup$
– Stockfish
Jan 7 at 11:58
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@Stockfish $ninmathcal{N}$
$endgroup$
– Riccardo.Alestra
Jan 7 at 12:12
$begingroup$
See page $64$ of my notes, or just square everything and apply induction on $n$.
$endgroup$
– Jack D'Aurizio
Jan 7 at 12:49
add a comment |
$begingroup$
How can I prove the following inequality?
$$sqrt{n+dfrac{1}{4}}ledfrac{n!}{left(n-dfrac{1}{2}right)!}lesqrt{n+dfrac{1}{4}+dfrac{1}{16n+12}}$$
Thanks
inequality
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
$endgroup$
How can I prove the following inequality?
$$sqrt{n+dfrac{1}{4}}ledfrac{n!}{left(n-dfrac{1}{2}right)!}lesqrt{n+dfrac{1}{4}+dfrac{1}{16n+12}}$$
Thanks
inequality
inequality
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
asked Jan 7 at 11:43
Riccardo.AlestraRiccardo.Alestra
6,01422354
6,01422354
closed as off-topic by A. Pongrácz, Chris Custer, José Carlos Santos, Cesareo, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – A. Pongrácz, Chris Custer, José Carlos Santos, Cesareo, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by A. Pongrácz, Chris Custer, José Carlos Santos, Cesareo, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – A. Pongrácz, Chris Custer, José Carlos Santos, Cesareo, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
$n in mathbb{N}$? What is $(n-1/2)!$ then?
$endgroup$
– Stockfish
Jan 7 at 11:58
$begingroup$
@Stockfish $ninmathcal{N}$
$endgroup$
– Riccardo.Alestra
Jan 7 at 12:12
$begingroup$
See page $64$ of my notes, or just square everything and apply induction on $n$.
$endgroup$
– Jack D'Aurizio
Jan 7 at 12:49
add a comment |
$begingroup$
$n in mathbb{N}$? What is $(n-1/2)!$ then?
$endgroup$
– Stockfish
Jan 7 at 11:58
$begingroup$
@Stockfish $ninmathcal{N}$
$endgroup$
– Riccardo.Alestra
Jan 7 at 12:12
$begingroup$
See page $64$ of my notes, or just square everything and apply induction on $n$.
$endgroup$
– Jack D'Aurizio
Jan 7 at 12:49
$begingroup$
$n in mathbb{N}$? What is $(n-1/2)!$ then?
$endgroup$
– Stockfish
Jan 7 at 11:58
$begingroup$
$n in mathbb{N}$? What is $(n-1/2)!$ then?
$endgroup$
– Stockfish
Jan 7 at 11:58
$begingroup$
@Stockfish $ninmathcal{N}$
$endgroup$
– Riccardo.Alestra
Jan 7 at 12:12
$begingroup$
@Stockfish $ninmathcal{N}$
$endgroup$
– Riccardo.Alestra
Jan 7 at 12:12
$begingroup$
See page $64$ of my notes, or just square everything and apply induction on $n$.
$endgroup$
– Jack D'Aurizio
Jan 7 at 12:49
$begingroup$
See page $64$ of my notes, or just square everything and apply induction on $n$.
$endgroup$
– Jack D'Aurizio
Jan 7 at 12:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is an idea that might lead to a solution.
If $z! = Gamma(z+1)$, then $left(n-dfrac{1}{2}right)! = Gamma left({tfrac {1}{2}}+nright)$ and we can use
$
displaystyle
Gamma left({tfrac {1}{2}}+nright)={(2n)! over 4^{n}n!}{sqrt {pi }}
$
(see Wikipedia) to get
$$
dfrac{n!}{left(n-dfrac{1}{2}right)!}
=
dfrac{4^nsqrt {pi }}{binom{2n}{n}}
$$
and use estimates for the central binomial coefficient. The ones given by
Wikipedia
$$
{frac {4^{n}}{{sqrt {4n}}}}leq {2n choose n}leq {frac {4^{n}}{{sqrt {3n+1}}}}
$$
are close but are not exactly what you want.
answered Jan 7 at 12:12
lhflhf
168k11172404
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an idea that might lead to a solution.
If $z! = Gamma(z+1)$, then $left(n-dfrac{1}{2}right)! = Gamma left({tfrac {1}{2}}+nright)$ and we can use
$
displaystyle
Gamma left({tfrac {1}{2}}+nright)={(2n)! over 4^{n}n!}{sqrt {pi }}
$
(see Wikipedia) to get
$$
dfrac{n!}{left(n-dfrac{1}{2}right)!}
=
dfrac{4^nsqrt {pi }}{binom{2n}{n}}
$$
and use estimates for the central binomial coefficient. The ones given by
Wikipedia
$$
{frac {4^{n}}{{sqrt {4n}}}}leq {2n choose n}leq {frac {4^{n}}{{sqrt {3n+1}}}}
$$
are close but are not exactly what you want.
answered Jan 7 at 12:12
lhflhf
168k11172404
$endgroup$
add a comment |
$begingroup$
Here is an idea that might lead to a solution.
If $z! = Gamma(z+1)$, then $left(n-dfrac{1}{2}right)! = Gamma left({tfrac {1}{2}}+nright)$ and we can use
$
displaystyle
Gamma left({tfrac {1}{2}}+nright)={(2n)! over 4^{n}n!}{sqrt {pi }}
$
(see Wikipedia) to get
$$
dfrac{n!}{left(n-dfrac{1}{2}right)!}
=
dfrac{4^nsqrt {pi }}{binom{2n}{n}}
$$
and use estimates for the central binomial coefficient. The ones given by
Wikipedia
$$
{frac {4^{n}}{{sqrt {4n}}}}leq {2n choose n}leq {frac {4^{n}}{{sqrt {3n+1}}}}
$$
are close but are not exactly what you want.
answered Jan 7 at 12:12
lhflhf
168k11172404
$endgroup$
add a comment |
$begingroup$
Here is an idea that might lead to a solution.
If $z! = Gamma(z+1)$, then $left(n-dfrac{1}{2}right)! = Gamma left({tfrac {1}{2}}+nright)$ and we can use
$
displaystyle
Gamma left({tfrac {1}{2}}+nright)={(2n)! over 4^{n}n!}{sqrt {pi }}
$
(see Wikipedia) to get
$$
dfrac{n!}{left(n-dfrac{1}{2}right)!}
=
dfrac{4^nsqrt {pi }}{binom{2n}{n}}
$$
and use estimates for the central binomial coefficient. The ones given by
Wikipedia
$$
{frac {4^{n}}{{sqrt {4n}}}}leq {2n choose n}leq {frac {4^{n}}{{sqrt {3n+1}}}}
$$
are close but are not exactly what you want.
answered Jan 7 at 12:12
lhflhf
168k11172404
$endgroup$
Here is an idea that might lead to a solution.
If $z! = Gamma(z+1)$, then $left(n-dfrac{1}{2}right)! = Gamma left({tfrac {1}{2}}+nright)$ and we can use
$
displaystyle
Gamma left({tfrac {1}{2}}+nright)={(2n)! over 4^{n}n!}{sqrt {pi }}
$
(see Wikipedia) to get
$$
dfrac{n!}{left(n-dfrac{1}{2}right)!}
=
dfrac{4^nsqrt {pi }}{binom{2n}{n}}
$$
and use estimates for the central binomial coefficient. The ones given by
Wikipedia
$$
{frac {4^{n}}{{sqrt {4n}}}}leq {2n choose n}leq {frac {4^{n}}{{sqrt {3n+1}}}}
$$
are close but are not exactly what you want.
answered Jan 7 at 12:12
lhflhf
168k11172404
answered Jan 7 at 12:12
lhflhf
168k11172404
answered Jan 7 at 12:12
lhflhf
168k11172404
answered Jan 7 at 12:12
lhflhf
168k11172404
168k11172404
add a comment |
add a comment |
$begingroup$
$n in mathbb{N}$? What is $(n-1/2)!$ then?
$endgroup$
– Stockfish
Jan 7 at 11:58
$begingroup$
@Stockfish $ninmathcal{N}$
$endgroup$
– Riccardo.Alestra
Jan 7 at 12:12
$begingroup$
See page $64$ of my notes, or just square everything and apply induction on $n$.
$endgroup$
– Jack D'Aurizio
Jan 7 at 12:49