Prove that $sqrt{frac{n-1}2} frac{Gammaleft[frac{n-1}2right]}{Gammaleft[frac{n}2right]}gt1quadforall n ge...












1












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How to prove




$$sqrt{frac{n-1}{2}} frac{Gammaleft[frac{n-1}{2}right]}{Gammaleft[frac{n}{2}right]} gt 1 quad forall n ge 2,nin mathbb{N}$$




I plot this function in Mathematica and verify it indeed is greater than $1$. I just know about some basic property of Gamma function.

Any hint? Thanks in advance!






gamma-function







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  • $begingroup$
    are the square brackets indicating floor ?
    $endgroup$
    – G Cab
    Jan 10 at 9:52










  • $begingroup$
    @GCab No,it has no other meaning.
    $endgroup$
    – Spaceship222
    Jan 10 at 10:01
















1












$begingroup$


How to prove




$$sqrt{frac{n-1}{2}} frac{Gammaleft[frac{n-1}{2}right]}{Gammaleft[frac{n}{2}right]} gt 1 quad forall n ge 2,nin mathbb{N}$$




I plot this function in Mathematica and verify it indeed is greater than $1$. I just know about some basic property of Gamma function.

Any hint? Thanks in advance!






gamma-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    are the square brackets indicating floor ?
    $endgroup$
    – G Cab
    Jan 10 at 9:52










  • $begingroup$
    @GCab No,it has no other meaning.
    $endgroup$
    – Spaceship222
    Jan 10 at 10:01














1












1








1





$begingroup$


How to prove




$$sqrt{frac{n-1}{2}} frac{Gammaleft[frac{n-1}{2}right]}{Gammaleft[frac{n}{2}right]} gt 1 quad forall n ge 2,nin mathbb{N}$$




I plot this function in Mathematica and verify it indeed is greater than $1$. I just know about some basic property of Gamma function.

Any hint? Thanks in advance!






gamma-function







share|cite|improve this question











$endgroup$




How to prove




$$sqrt{frac{n-1}{2}} frac{Gammaleft[frac{n-1}{2}right]}{Gammaleft[frac{n}{2}right]} gt 1 quad forall n ge 2,nin mathbb{N}$$




I plot this function in Mathematica and verify it indeed is greater than $1$. I just know about some basic property of Gamma function.

Any hint? Thanks in advance!






gamma-function




gamma-function






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share|cite|improve this question













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edited Jan 10 at 8:43









mrtaurho

4,44121235




4,44121235










asked Jan 10 at 7:01









Spaceship222Spaceship222

6217




6217












  • $begingroup$
    are the square brackets indicating floor ?
    $endgroup$
    – G Cab
    Jan 10 at 9:52










  • $begingroup$
    @GCab No,it has no other meaning.
    $endgroup$
    – Spaceship222
    Jan 10 at 10:01


















  • $begingroup$
    are the square brackets indicating floor ?
    $endgroup$
    – G Cab
    Jan 10 at 9:52










  • $begingroup$
    @GCab No,it has no other meaning.
    $endgroup$
    – Spaceship222
    Jan 10 at 10:01
















$begingroup$
are the square brackets indicating floor ?
$endgroup$
– G Cab
Jan 10 at 9:52




$begingroup$
are the square brackets indicating floor ?
$endgroup$
– G Cab
Jan 10 at 9:52












$begingroup$
@GCab No,it has no other meaning.
$endgroup$
– Spaceship222
Jan 10 at 10:01




$begingroup$
@GCab No,it has no other meaning.
$endgroup$
– Spaceship222
Jan 10 at 10:01










2 Answers
2






active

oldest

votes


















0












$begingroup$

Putting
$$
F(n) = {{sqrt {{{n - 1} over 2}} Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)}}
$$
then
$$
eqalign{
& F(n)^{,2} = {{left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)}
over {Gamma left( {{n over 2}} right)^{,2} }} =
{{Gamma left( {{{n + 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)^{,2} }} cr}
$$
and the result follows from the log-convexity of Gamma
$$
ln F(n) = {1 over 2}left( {ln Gamma left( {{{n + 1} over 2}} right) + ln Gamma left( {{{n - 1} over 2}} right)} right)
- ln Gamma left( {{n over 2}} right) > 0quad left| {;1 < n in mathbb R} right.
$$






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$endgroup$













  • $begingroup$
    Nice proof!Thanks!
    $endgroup$
    – Spaceship222
    Jan 10 at 14:24










  • $begingroup$
    @Spaceship222: glad for the appreciation !
    $endgroup$
    – G Cab
    Jan 10 at 16:48



















0












$begingroup$

At least for "large" values of $n$.



Let $$y=sqrt{frac{n-1}{2}},, frac{Gamma[frac{n-1}{2}]}{Gamma[frac{n}{2}]} $$
$$log(y)=frac 12 log(n-1)-frac 12 log(2) + log left(Gamma left(frac{n-1}{2}right)right)-log left(Gamma left(frac{n}{2}right)right)$$ Now, use Stirling approximation
$$log(Gamma(p))=p (log (p)-1)+frac{1}{2} left(-log left({p}right)+log (2 pi
)right)+frac{1}{12 p}-frac{1}{360
p^3}+Oleft(frac{1}{p^5}right)$$ and get (continuing with Taylor expansions)
$$log(y)=frac{1}{4 n}+frac{1}{4 n^2}+frac{5}{24
n^3}+Oleft(frac{1}{n^4}right)$$
$$y=e^{log(y)}=1+frac{1}{4 n}+frac{9}{32 n^2}+frac{35}{128 n^3}+Oleft(frac{1}{n^4}right)$$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Putting
    $$
    F(n) = {{sqrt {{{n - 1} over 2}} Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)}}
    $$
    then
    $$
    eqalign{
    & F(n)^{,2} = {{left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)}
    over {Gamma left( {{n over 2}} right)^{,2} }} =
    {{Gamma left( {{{n + 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)^{,2} }} cr}
    $$
    and the result follows from the log-convexity of Gamma
    $$
    ln F(n) = {1 over 2}left( {ln Gamma left( {{{n + 1} over 2}} right) + ln Gamma left( {{{n - 1} over 2}} right)} right)
    - ln Gamma left( {{n over 2}} right) > 0quad left| {;1 < n in mathbb R} right.
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice proof!Thanks!
      $endgroup$
      – Spaceship222
      Jan 10 at 14:24










    • $begingroup$
      @Spaceship222: glad for the appreciation !
      $endgroup$
      – G Cab
      Jan 10 at 16:48
















    0












    $begingroup$

    Putting
    $$
    F(n) = {{sqrt {{{n - 1} over 2}} Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)}}
    $$
    then
    $$
    eqalign{
    & F(n)^{,2} = {{left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)}
    over {Gamma left( {{n over 2}} right)^{,2} }} =
    {{Gamma left( {{{n + 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)^{,2} }} cr}
    $$
    and the result follows from the log-convexity of Gamma
    $$
    ln F(n) = {1 over 2}left( {ln Gamma left( {{{n + 1} over 2}} right) + ln Gamma left( {{{n - 1} over 2}} right)} right)
    - ln Gamma left( {{n over 2}} right) > 0quad left| {;1 < n in mathbb R} right.
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice proof!Thanks!
      $endgroup$
      – Spaceship222
      Jan 10 at 14:24










    • $begingroup$
      @Spaceship222: glad for the appreciation !
      $endgroup$
      – G Cab
      Jan 10 at 16:48














    0












    0








    0





    $begingroup$

    Putting
    $$
    F(n) = {{sqrt {{{n - 1} over 2}} Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)}}
    $$
    then
    $$
    eqalign{
    & F(n)^{,2} = {{left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)}
    over {Gamma left( {{n over 2}} right)^{,2} }} =
    {{Gamma left( {{{n + 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)^{,2} }} cr}
    $$
    and the result follows from the log-convexity of Gamma
    $$
    ln F(n) = {1 over 2}left( {ln Gamma left( {{{n + 1} over 2}} right) + ln Gamma left( {{{n - 1} over 2}} right)} right)
    - ln Gamma left( {{n over 2}} right) > 0quad left| {;1 < n in mathbb R} right.
    $$






    share|cite|improve this answer











    $endgroup$



    Putting
    $$
    F(n) = {{sqrt {{{n - 1} over 2}} Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)}}
    $$
    then
    $$
    eqalign{
    & F(n)^{,2} = {{left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)}
    over {Gamma left( {{n over 2}} right)^{,2} }} =
    {{Gamma left( {{{n + 1} over 2}} right)Gamma left( {{{n - 1} over 2}} right)} over {Gamma left( {{n over 2}} right)^{,2} }} cr}
    $$
    and the result follows from the log-convexity of Gamma
    $$
    ln F(n) = {1 over 2}left( {ln Gamma left( {{{n + 1} over 2}} right) + ln Gamma left( {{{n - 1} over 2}} right)} right)
    - ln Gamma left( {{n over 2}} right) > 0quad left| {;1 < n in mathbb R} right.
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 10 at 16:59

























    answered Jan 10 at 10:56









    G CabG Cab

    19k31238




    19k31238












    • $begingroup$
      Nice proof!Thanks!
      $endgroup$
      – Spaceship222
      Jan 10 at 14:24










    • $begingroup$
      @Spaceship222: glad for the appreciation !
      $endgroup$
      – G Cab
      Jan 10 at 16:48


















    • $begingroup$
      Nice proof!Thanks!
      $endgroup$
      – Spaceship222
      Jan 10 at 14:24










    • $begingroup$
      @Spaceship222: glad for the appreciation !
      $endgroup$
      – G Cab
      Jan 10 at 16:48
















    $begingroup$
    Nice proof!Thanks!
    $endgroup$
    – Spaceship222
    Jan 10 at 14:24




    $begingroup$
    Nice proof!Thanks!
    $endgroup$
    – Spaceship222
    Jan 10 at 14:24












    $begingroup$
    @Spaceship222: glad for the appreciation !
    $endgroup$
    – G Cab
    Jan 10 at 16:48




    $begingroup$
    @Spaceship222: glad for the appreciation !
    $endgroup$
    – G Cab
    Jan 10 at 16:48











    0












    $begingroup$

    At least for "large" values of $n$.



    Let $$y=sqrt{frac{n-1}{2}},, frac{Gamma[frac{n-1}{2}]}{Gamma[frac{n}{2}]} $$
    $$log(y)=frac 12 log(n-1)-frac 12 log(2) + log left(Gamma left(frac{n-1}{2}right)right)-log left(Gamma left(frac{n}{2}right)right)$$ Now, use Stirling approximation
    $$log(Gamma(p))=p (log (p)-1)+frac{1}{2} left(-log left({p}right)+log (2 pi
    )right)+frac{1}{12 p}-frac{1}{360
    p^3}+Oleft(frac{1}{p^5}right)$$     and get (continuing with Taylor expansions)
    $$log(y)=frac{1}{4 n}+frac{1}{4 n^2}+frac{5}{24
    n^3}+Oleft(frac{1}{n^4}right)$$
    $$y=e^{log(y)}=1+frac{1}{4 n}+frac{9}{32 n^2}+frac{35}{128 n^3}+Oleft(frac{1}{n^4}right)$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      At least for "large" values of $n$.



      Let $$y=sqrt{frac{n-1}{2}},, frac{Gamma[frac{n-1}{2}]}{Gamma[frac{n}{2}]} $$
      $$log(y)=frac 12 log(n-1)-frac 12 log(2) + log left(Gamma left(frac{n-1}{2}right)right)-log left(Gamma left(frac{n}{2}right)right)$$ Now, use Stirling approximation
      $$log(Gamma(p))=p (log (p)-1)+frac{1}{2} left(-log left({p}right)+log (2 pi
      )right)+frac{1}{12 p}-frac{1}{360
      p^3}+Oleft(frac{1}{p^5}right)$$       and get (continuing with Taylor expansions)
      $$log(y)=frac{1}{4 n}+frac{1}{4 n^2}+frac{5}{24
      n^3}+Oleft(frac{1}{n^4}right)$$
      $$y=e^{log(y)}=1+frac{1}{4 n}+frac{9}{32 n^2}+frac{35}{128 n^3}+Oleft(frac{1}{n^4}right)$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        At least for "large" values of $n$.



        Let $$y=sqrt{frac{n-1}{2}},, frac{Gamma[frac{n-1}{2}]}{Gamma[frac{n}{2}]} $$
        $$log(y)=frac 12 log(n-1)-frac 12 log(2) + log left(Gamma left(frac{n-1}{2}right)right)-log left(Gamma left(frac{n}{2}right)right)$$ Now, use Stirling approximation
        $$log(Gamma(p))=p (log (p)-1)+frac{1}{2} left(-log left({p}right)+log (2 pi
        )right)+frac{1}{12 p}-frac{1}{360
        p^3}+Oleft(frac{1}{p^5}right)$$         and get (continuing with Taylor expansions)
        $$log(y)=frac{1}{4 n}+frac{1}{4 n^2}+frac{5}{24
        n^3}+Oleft(frac{1}{n^4}right)$$
        $$y=e^{log(y)}=1+frac{1}{4 n}+frac{9}{32 n^2}+frac{35}{128 n^3}+Oleft(frac{1}{n^4}right)$$






        share|cite|improve this answer









        $endgroup$



        At least for "large" values of $n$.



        Let $$y=sqrt{frac{n-1}{2}},, frac{Gamma[frac{n-1}{2}]}{Gamma[frac{n}{2}]} $$
        $$log(y)=frac 12 log(n-1)-frac 12 log(2) + log left(Gamma left(frac{n-1}{2}right)right)-log left(Gamma left(frac{n}{2}right)right)$$ Now, use Stirling approximation
        $$log(Gamma(p))=p (log (p)-1)+frac{1}{2} left(-log left({p}right)+log (2 pi
        )right)+frac{1}{12 p}-frac{1}{360
        p^3}+Oleft(frac{1}{p^5}right)$$         and get (continuing with Taylor expansions)
        $$log(y)=frac{1}{4 n}+frac{1}{4 n^2}+frac{5}{24
        n^3}+Oleft(frac{1}{n^4}right)$$
        $$y=e^{log(y)}=1+frac{1}{4 n}+frac{9}{32 n^2}+frac{35}{128 n^3}+Oleft(frac{1}{n^4}right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 8:35









        Claude LeiboviciClaude Leibovici

        121k1157133




        121k1157133






























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