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Show that, if $n$ is a positive integer, $$\frac1{\sqrt{(n+\frac12) \pi}} \le\frac{1\cdot 3\cdot 5 ... (2n-1)}{2\cdot 4\cdot 6 ... (2n)} \le \frac1{\sqrt{n \pi}} . $$

This result is in a current issue of a MAA magazine, and I thought it would be interesting to see how many proofs could be found.

I know it may be a duplicate, but none were suggested when I entered it.

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  • $\begingroup$ This is related to the famous Wallis product for $\pi$, and Pedro Tamaroff's response below is based on Wallis' integrals. $\endgroup$ – Lucian Dec 5 '13 at 5:59
  • $\begingroup$ I'm curious to see the context in the magazine; is it the October issue of Math. Mag.? $\endgroup$ – user21467 Dec 7 '13 at 2:47
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    $\begingroup$ See also math.stackexchange.com/questions/58560/… $\endgroup$ – user940 Dec 7 '13 at 3:15
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    $\begingroup$ @StevenTaschuk "Inequalities for Gamma Function Ratios", G.J.O. Jameson American Mathematical Monthly, December 2013, pp 936-940. $\endgroup$ – user940 Dec 7 '13 at 13:41
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Since $0\leqslant \sin t\leqslant 1 $ over $[0,\pi/2]$ we have that $$\int_0^{\pi/2}\sin^{2n+1}t dt\leqslant \int_0^{\pi/2}\sin^{2n}tdt\leqslant \int_0^{\pi/2}\sin^{2n-1}t dt$$

Now, it is not hard to show that if $I_k= \displaystyle\int_0^{\pi/2}\sin^{k}t$ then $I_k=\dfrac{k-1}{k}I_{k-2}$, by integrating by parts. Using $I_1=1$ and $I_0=\dfrac{\pi}2$, we get that

$$\int_0^{\pi/2}\sin^{m}tdt=\begin{cases}\dfrac{4^n}{2n+1}\dbinom{2n}n^{-1},&m=2n+1\text{ odd}\\\dfrac 1 {4^n}\dbinom{2n}n\dfrac \pi 2,&m=2n\text{ even}\end{cases}$$

Substituting in the above, one gets

$$2n\left(\frac 1 {4^n}\binom {2n}n\right)^2\leqslant \frac{2}\pi\leqslant (2n+1)\left(\frac 1 {4^n}\binom {2n}n\right)^2$$

which becomes$$ \frac{1}{{2n}}{\left( {\frac{1}{{{4^n}}}\binom{{2n}}{n}} \right)^{ - 2}} \geqslant \frac{\pi }{2} \geqslant \frac{1}{{2n + 1}}{\left( {\frac{1}{{{4^n}}}\binom{{2n}}{n}} \right)^{ - 2}} $$ and in turn gives $$ \frac{1}{{\pi n}} \geqslant {\left( {\frac{1}{{{4^n}}}\binom{{2n}}{n}} \right)^2} \geqslant \frac{1}{\pi }\frac{1}{{n + 1/2}} $$

As desired.

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    $\begingroup$ @user17762 PEW PEW PEW. $\endgroup$ – Pedro Tamaroff Dec 5 '13 at 4:44
  • $\begingroup$ Add the details and I will accept. $\endgroup$ – marty cohen Dec 7 '13 at 1:57
  • $\begingroup$ Your final inequality got turned around. $\endgroup$ – user940 Dec 7 '13 at 13:23
  • $\begingroup$ @ByronSchmuland The risks of copy-pasting the previous line. =) $\endgroup$ – Pedro Tamaroff Dec 7 '13 at 15:00

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