Show that $$\frac{\sqrt{n-1}\Gamma((n-1)/2)}{\sqrt{2}\Gamma(n/2)}>1$$

I tried to solve it using Taylor series expansion.


In this answer, it is shown that $\Gamma(x)$ is log-convex. Therefore, $$ \begin{align} \Gamma\left(\frac n2\right) &\le\Gamma\left(\frac{n-1}2\right)^{1/2}\Gamma\left(\frac{n+1}2\right)^{1/2}\\[6pt] &=\sqrt{\frac{n-1}2}\ \Gamma\left(\frac{n-1}2\right) \end{align} $$ since $\Gamma(x+1)=x\Gamma(x)$.

Implication of Log-Convexity

A log-convex function is a function whose logarithm is convex. Thus, if $f$ is log-convex, we have $$ \log\circ f\left(\frac{x+y}2\right)\le\frac{\log\circ f(x)+\log\circ f(y)}2 $$ removing the $\log$s by applying $\exp$, which is monotonically increasing, we get $$ f\left(\frac{x+y}2\right)\le f(x)^{1/2}f(y)^{1/2} $$

  • $\begingroup$ Would the downvoter care to comment? $\endgroup$ – robjohn May 5 '14 at 16:43
  • $\begingroup$ Thanks, I get what a log-convex function is. Unfortunately, I don't understand how the fact that \Gamma(x) is log-convex yields to the inequality above. Could you please explain it to me a little bit? $\endgroup$ – Filip May 15 '14 at 15:40
  • $\begingroup$ @Filip: I've added a section on log-convexity. Let me know if there is still confusion. $\endgroup$ – robjohn May 15 '14 at 16:29
  • $\begingroup$ Cool, thanks a lot! $\endgroup$ – Filip May 15 '14 at 16:39

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