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.

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 $n\geq 1$.
