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I would like to prove $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)} \le \frac{1}{\sqrt{n}}$$ for all natural $n \ge 1$. The inequality does seem to be true numerically, but the proof eludes me.

  • 1
    [Stirling](http://mathworld.wolfram.com/StirlingsApproximation.html) would be of great service...2010-12-22
  • 0
    Have you tried a proof by induction? Do you know the value of $\Gamma(\pi)$?2010-12-22
  • 0
    @cch: Yes, and yes.2010-12-22
  • 1
    On reconsideration: Yes, and no.2010-12-22
  • 0
    You might find this useful: http://math.stackexchange.com/questions/9372/why-doesnt-this-series-converge/9379#93792010-12-22
  • 0
    Sorry -- I meant to type $\Gamma(1/2)$ above -- you would need that for an induction proof.2010-12-22
  • 1
    @J. M.: Maybe looking at some specific proof of Stirling's formula would help - but to me that is an asymptotic formula.2010-12-22

7 Answers 7

-1

\begin{eqnarray*} \Gamma(x+1)>\frac{\frac{(900\gamma^2+73\pi^2)x}{900\gamma}+\frac{73}{100}}{\frac{\pi^2 x}{9\gamma}+1}\quad \textbf{[1]} \end{eqnarray*} i want to prove this inquality on $(0,1) is increasing i defined it as

i defined it as

\begin{eqnarray*} w(x)=\frac{\log[\Gamma(x+1)(\frac{\pi^2 x}{9\gamma}+1)]}{\log[\frac{(900\gamma^2+73\pi^2)x}{900\gamma}+\frac{73}{100}]} \quad [2] \end{eqnarray*} but [2] not increasing to prove it ! what it need to get it increasing

12

Here's a direct proof:

The gamma function satisfies $\Gamma(n+1) = n \Gamma(n)$, so the function $g(x) = \ln \Gamma(x)$ satisfies $g(n+1) = \ln n + g(n)$. In other words, the line segment between the points $(n,g(n))$ and $(n+1,g(n+1))$ has slope $\ln n$.

Moreover, $g$ is known to be a convex function for $x>0$ (cf. the Bohr–Mollerup theorem), so the point $(n+1/2, g(n+1/2))$ lies below that line segment: $g(n+1/2) < g(n+1) - \frac12 \ln n$. Exponentiating both sides gives $\Gamma(n+1/2) < \Gamma(n+1) / \sqrt{n}$, as desired.

  • 1
    I suspect that Gautschi's inequality is proved in a similar way. Incidentally, $n$ doesn't have to be an integer.2010-12-22
  • 0
    For a proof that $\Gamma$ is log-convex (using Hölder's inequality), see PlanetMath: http://planetmath.org/?op=getobj&from=objects&id=3808.2010-12-24
  • 0
    Here is [a proof of the log-convexity of $\Gamma$](http://math.stackexchange.com/a/101007/) using Young's Inequality.2012-01-27
7

For a completely elementary proof let

$$I_n= \int_0^{ \pi/2} \sin^n x \textrm{ d}x $$

then integration by parts gives

$$I_n = \frac{n-1}{n} I_{n-2} . \quad (1) $$

From which we get

$$I_{2n}= \frac{2n-1}{2n} \frac{2n-3}{2n-2} \cdots \frac{1}{2} \frac{\pi}{2} \quad (2) $$

and

$$I_{2n+1}= \frac{2n}{2n+1} \frac{2n-2}{2n-1} \cdots \frac{2}{3} . \quad (3) $$

Since $ \sin^{2n+1} x < \sin^{2n} x $ for $ x \in (0,\pi/2) $ we have

$$I_{2n+1}< I_{2n} . \quad (4) $$

Also, from $(1)$ we have $(2n+1)I_{2n+1} = 2nI_{2n-1} > 2nI_{2n}$ since $I_{2n-1} >I_{2n}.$

Therefore using this and $(4)$

$$ \frac{2n+1}{2n} > \frac{ I_{2n} }{ I_{2n+1} } > 1 . \quad (5) $$

But from $(2)$ and $(3)$

$$ \frac{ I_{2n} }{ I_{2n+1} } = \frac{2n+1}{2} \frac{ (2n!)^2 }{ 4^{2n} (n!)^4 } \pi .$$

Putting this in $(5)$ gives

$$ \frac{1}{ \sqrt{n} } > \frac{1}{4^n} { 2n \choose n } \sqrt{ \pi } > \frac{1}{ \sqrt{ n + 1/2} }.$$

i.e.

$$ \frac{1}{ \sqrt{n} } > \frac{ \Gamma(n + 1/2) }{ \Gamma(n+1) } > \frac{1}{ \sqrt{ n + 1/2} }.$$

  • 1
    Tsk, I was expecting that it would be Moron who would post this proof... :D2010-12-23
  • 0
    @J.M. Yes, one develops a feel for the sort of mathematics that our members enjoy.2010-12-23
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    @J.M: Wallis! Wallis! Wallis! :-)2010-12-23
6

This follows directly from Gautschi's inequality, valid for $0 < s < 1$ and $x > 0$:

$$x^{1-s} < \frac{\Gamma(x+1)}{\Gamma(x+s)} < (x+1)^{1-s}.$$

As a side note, the link is to NIST's new (as of this year) Digital Library of Mathematical Functions. The DLMF is an update of the classic Abramowitz and Stegun text, Handbook of Mathematical Functions.

  • 0
    Mike, I hope you don't mind if I changed the link to link directly to the formula. In any event, DLMF gives [this](http://www.ams.org/mathscinet-getitem?mr=0103289) as the reference for Gautschi's inequality; sadly there seems to be no digital archive for the *Journal of Mathematics and Physics*.2010-12-23
  • 0
    @J.M.: That's fine.2010-12-23
6

To elaborate on Shai Covo's answer:

$$2n\left[{2n\choose n} 4^{-n}\right]^2 = {1\over2}\,{3\over2}\,{3\over4}\,{5\over4}\cdots {{2n-1}\over{2n-2}}\,{{2n-1}\over{2n}} ={1\over 2}\prod_{j=2}^n\left(1+{1\over4j(j-1)}\right). $$

By Wallis's formula, the middle expression converges to $2\over\pi$. The right hand expression shows that it is strictly increasing. Therefore, multiplying by $\pi/2$ and taking square roots shows that
$$\sqrt{n\pi}{2n\choose n} 4^{-n}\uparrow 1.$$

3

Since $$ \Gamma \bigg(n + \frac{1}{2}\bigg) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi $$ (see e.g. here), we have $$ \frac{{\Gamma (n + \frac{1}{2})}}{{\Gamma (n+1)}} = \frac{{{2n \choose n}}}{{4^n }}\sqrt \pi . $$ This might be very useful, but I have to check.

  • 0
    Well, at least Byron found this useful...2010-12-23
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    Yes, so I'll vote it up.2010-12-23
1

I've come up with yet another proof:
Lemma. Let $0<\alpha Proof: $$ \Gamma(z)=\int_0^{\infty} t^{z-1}e^{-t}dt=\int_0^{\infty} t^{\frac{(z-\alpha)-1}{2}} t^{\frac{(z+\alpha)-1}{2}} e^{-t}dt=\langle\cdot^{\frac{(z-\alpha)-1}{2}},\;\cdot^{\frac{(z+\alpha)-1}{2}} \rangle,\\ $$ where $\langle f,\;g \rangle:=\int_0^{\infty} f(t)g(t)e^{-t}dt$. Now apply Cauchy–Schwarz inequality and since functions $\cdot^{\frac{(z-\alpha)-1}{2}}\;\;\text{and}\;\;\cdot^{\frac{(z+\alpha)-1}{2}}$ are linearly independent, this inequality is strict: $$ \langle\cdot^{\frac{(z-\alpha)-1}{2}},\;\cdot^{\frac{(z+\alpha)-1}{2}} \rangle<\left(\int_0^{\infty} t^{(z-\alpha)-1} e^{-t}dt\cdot\int_0^{\infty} t^{(z+\alpha)-1} e^{-t}dt\right)^{1/2}=\sqrt{\Gamma(z-\alpha)\Gamma(z+\alpha)}.\;\;\blacksquare $$ Now take $x>0$ and apply the aforementioned lemma with $\alpha=\frac{1}{2}$: $$ \frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma(x+1)}<\frac{\sqrt{\Gamma(x)\Gamma(x+1)}}{\Gamma(x+1)}=\sqrt{\frac{\Gamma(x)}{x\Gamma(x)}}=\frac{1}{\sqrt{x}}.\;\;\;\blacksquare $$