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According to WolframAlpha, the following sequence \begin{align*} a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})} \end{align*} seems to converges to zero. However, how can I prove this ? The difficulty I encounter is this "one quarter of an integer" in the Gamma functions.

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  • $\begingroup$ The magic words are: Stirling's formula. (check out the wikipedia article on the gamma function). $\endgroup$ – Igor Rivin Mar 2 at 18:10
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    $\begingroup$ Just for your interest: $$ a_n \sim \sqrt {\frac{2}{n}} \left( {1 - \frac{1}{{16n^2 }} + \frac{{21}}{{512n^4 }} - \cdots } \right) $$ as $n\to +\infty$. You can derive it from dlmf.nist.gov/5.11.E14 $\endgroup$ – Gary Mar 2 at 19:00
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    $\begingroup$ Follows from the limit which is easy to remember: for any $a$ $$\lim_{x\to\infty}\frac{\Gamma(x+a)}{x^a\,\Gamma(x)}=1.$$ $\endgroup$ – metamorphy Mar 3 at 0:47
  • $\begingroup$ @ metamorphy this is related to the Pochhammer symbol $(x)_a=\frac{\Gamma(x+a)}{\Gamma(x)}$ which can be evaluated heuristically as $ x(x+1)(x+2)...(x+a-1)\underset{x \to \infty}\simeq x^a$ $\endgroup$ – Dr. Wolfgang Hintze Mar 3 at 8:28
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First approach. By the beta integral \begin{align*} \frac{{\Gamma (x)}}{{\Gamma \left( {x + \tfrac{1}{2}} \right)}} & = \frac{1}{{\Gamma \left( {\frac{1}{2}} \right)}}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}\int_0^1 {\frac{{t^{x - 1} }}{{\sqrt {1 - t} }}dt} \\ & = \frac{1}{{\sqrt \pi }}\int_0^{ + \infty } {\frac{{e^{ - xs} }}{{\sqrt {1 - e^{ - s} } }}ds} = \frac{1}{{\sqrt \pi }}\int_0^{ + \infty } {e^{ - (x - 1/4)s} s^{ - 1/2} \sqrt {\frac{{s/2}}{{\sinh (s/2)}}} ds} \end{align*} for all $x>0$. It is well known that $w < \sinh w$ for all $w>0$. Thus, for $x>\frac{1}{4}$, $$ \!\! \frac{{\Gamma (x)}}{{\Gamma \left( {x + \frac{1}{2}} \right)}} \le \frac{1}{{\sqrt \pi }}\int_0^{ + \infty }\! {e^{ - (x - 1/4)s} s^{ - 1/2} ds} = \frac{1}{{\sqrt {\pi \!\left( {x - \frac{1}{4}} \right)} }}\int_0^{ + \infty }\! {e^{ - t} t^{ - 1/2} dt} = \frac{1}{{\sqrt {x - \frac{1}{4}} }}. $$ Consequently, $$ \frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \sqrt {\frac{2}{n}} $$ for all $n\geq 1$. From the asymptotics I gave in the comments, it is seen that this bound is sharp.

Second approach. By the log-convexity and the functional equation of the gamma function, we have $$ \log \Gamma \left( {x + \tfrac{1}{2}} \right) \le \frac{1}{2}\log \Gamma (x) + \frac{1}{2}\log \Gamma (x + 1) = \log \Gamma (x + 1) - \frac{1}{2}\log x, $$ i.e., $$ \frac{{\Gamma \left( {x + \frac{1}{2}} \right)}}{{\Gamma (x + 1)}} \le \frac{1}{{\sqrt x }} $$ for all $x>0$. Consequently, $$ \frac{{\Gamma \left( {\frac{n}{2} + \frac{1}{4}} \right)}}{{\Gamma \left( {\frac{n}{2} + \frac{3}{4}} \right)}} \le \frac{2}{{\sqrt {2n - 1} }} $$ for all $n\geq 1$.

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Just to add a little bit to @Igor Rivin's answer - Stirling's formula gives you an asymptotic approximation for ${\Gamma(x)}$. That means that $$ \lim_{x\to\infty}\frac{\Gamma(x)}{S(x)} = 1 $$ (where ${S(x)}$ is the Stirling formula). We write ${\Gamma(x)\sim S(x)}$ to represent this fact (of course we are assuming ${S(x)\neq 0}$). You need to use the following proposition:

If ${f\sim g}$ and ${h\sim j}$, we have $$ \lim_{x\to\infty}\frac{f(x)}{h(x)} = \lim_{x\to \infty}\frac{g(x)}{j(x)} $$ (provided ${h(x)\neq 0}$ and ${j(x)\neq 0}$). It's easy to prove this by using properties of limits.

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Similar to Gary's answer but slightly shorter for the asymptotics.

With $x=\frac{n}{2}+\frac{1}{4}$ and $y=\frac{1}{2}$ the Beta integral becomes

$$\text{Beta}(x,y)=\int_{0}^{1} t^{x-1}(1-t)^{y-1}\,dt\\ \overset{t\to exp(-u/x)}=\frac{1}{x} \int_{0}^{\infty} e^{-u}(1-e^{-u/x})^{y-1}\,du\\ \overset{exp(-u/x)\simeq 1-\frac{u}{x}}\simeq\frac{1}{x} \int_{0}^{\infty} e^{-u}(u/x)^{y-1}\,du = x^{-y}\; \Gamma(y)=x^{-y}\sqrt{\pi}$$

whence follows that the original expression for $n \to \infty$ goes like $\frac{1}{\sqrt{x}}=\sqrt{\frac{2}{n}}$.

EDIT: The obvious (and easy to remember) substitution $t^x \to v$ gives the even shorter sequence for the leading term of the asymptotics:

$$\text{Beta}(x,y) = \frac{1}{x} \int_{0}^{1} (1-v^{\frac{1}{x}})^{y-1}\,dv=\frac{1}{x} \int_{0}^{1} (1-e^{\frac{\log(v)}{x}})^{y-1}\,dv\\ \simeq \frac{1}{x} \int_{0}^{1} (\frac{-\log(v)}{x})^{y-1}\,dv=x^{-y}\; \Gamma(y) $$ where Euler's favorite form of $\Gamma$ appears.

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$$a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}\implies \log(a_n)=\log \left(\Gamma \left(\frac{n}{2}+\frac{1}{4}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}+\frac{3}{4}\right)\right)$$ Now, using Stirling approximation $$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \log \left(\frac{2 \pi }{p}\right)+\frac{1}{12 p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ apply it twice and continue with Taylor expansion to obtain $$\log(a_n)=-\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{1}{16 n^2}+\frac{5}{128 n^4}+O\left(\frac{1}{n^5}\right)$$ Continuing with Taylor $$a_n=e^{\log(a_n)}=\sqrt{\frac{2}{n}} \exp\Big[-\frac{1}{16 n^2}+\frac{5}{128 n^4} \Big]$$

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