Showing a function grows like a certain power of $n$ Consider $f(n) = \frac{1}{n} {n \choose (n+1)/2} \frac{1}{2^n}$ for odd $n$. I suspect that $\sum_{n \geq 1} n^{\beta}f(n)= \infty$ for $\beta \geq 1/2$. For this purpose, it would be enough to show that $f(n) \sim n^{-3/2}$. How can I show this last statement?


Letting $n=2m+1, m\in\mathbb{N}$ we obtain
  \begin{align*}
\color{blue}{f(2m+1)}&=\frac{1}{2^{2m+1}(2m+1)}\binom{2m+1}{m+1}\\
&=\frac{1}{2^{2m+1}(m+1)}\binom{2m}{m}\tag{1}\\
&\,\,\color{blue}{=\frac{1}{2^{2m+1}}C_m}\tag{2}
\end{align*}
  with $C_m$ the $m$-th Catalan-number.

In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

The asymptotic behaviour of the Catalan numbers $C_m$ can be shown using Stirling's formula and is according to this MSE answer
\begin{align*}
C_m\sim\frac{4^m}{\sqrt{\pi}m^{\frac{3}{2}}}
\end{align*}
We conclude together with (2)
  \begin{align*}
\color{blue}{f(2m+1)\sim \frac{1}{2\sqrt{\pi}m^{\frac{3}{2}}}}
\end{align*}

Note: Since $f(n)$ is given for odd $n$ only, we do not know anything about the behaviour of $f$ for even arguments.
