A gamma function inequality

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.

• Stirling would be of great service... – J. M. is a poor mathematician Dec 22 '10 at 16:50
• Have you tried a proof by induction? Do you know the value of $\Gamma(\pi)$? – cch Dec 22 '10 at 17:05
• @cch: Yes, and yes. – heiner Dec 22 '10 at 17:23
• On reconsideration: Yes, and no. – heiner Dec 22 '10 at 17:31
• @J. M.: Maybe looking at some specific proof of Stirling's formula would help - but to me that is an asymptotic formula. – AD. Dec 22 '10 at 19:15

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.

• Mike, I hope you don't mind if I changed the link to link directly to the formula. In any event, DLMF gives this as the reference for Gautschi's inequality; sadly there seems to be no digital archive for the Journal of Mathematics and Physics. – J. M. is a poor mathematician Dec 23 '10 at 8:39
• @J.M.: That's fine. – Mike Spivey Dec 23 '10 at 16:44

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.

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.

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

• Tsk, I was expecting that it would be Moron who would post this proof... :D – J. M. is a poor mathematician Dec 23 '10 at 1:01
• @J.M. Yes, one develops a feel for the sort of mathematics that our members enjoy. – Derek Jennings Dec 23 '10 at 8:29
• @J.M: Wallis! Wallis! Wallis! :-) – Aryabhata Dec 23 '10 at 16:56

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
$$\sqrt{n\pi}{2n\choose n} 4^{-n}\uparrow 1.$$

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.

• Well, at least Byron found this useful... – Shai Covo Dec 23 '10 at 1:09
• Yes, so I'll vote it up. – user940 Dec 23 '10 at 3:10

I've come up with yet another proof:
Lemma. Let $0<\alpha<z$. Then $\Gamma(z)<\sqrt{\Gamma(z-\alpha)\Gamma(z+\alpha)}$.
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 $\alpha=\frac{1}{2}$: $$\frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma(x+1)}<\frac{\sqrt{\Gamma(x)\Gamma(x+1)}}{\Gamma(x+1)}=\sqrt{\frac{\Gamma(x)}{x\Gamma(x)}}=\frac{1}{\sqrt{x}}.\;\;\;\blacksquare$$

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

i defined it as

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