Show that $a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}$ converges to zero 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.
 A: 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 $n\geq 1$.
A: 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.
A: 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.
A: $$a_n = \frac{\Gamma(\frac{n}{2}+\frac{1}{4})}{\Gamma(\frac{n}{2}+\frac{3}{4})}\implies \log(a_n)=\log \left(\Gamma \left(\frac{n}{2}+\frac{1}{4}\right)\right)-\log \left(\Gamma \left(\frac{n}{2}+\frac{3}{4}\right)\right)$$ 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
$$a_n=e^{\log(a_n)}=\sqrt{\frac{2}{n}} \exp\Big[-\frac{1}{16 n^2}+\frac{5}{128
   n^4} \Big]$$
