Monotonicity of a sequence $\frac{(\mu/2)^n}{n!}K_{n-\frac{1}{2}}(\mu)$

Question

I want to investigate the monotonicity of the sequence $\{p_n\}$ given by \begin{equation} p_n=\frac{(\mu/2)^n}{n!}K_{n-\frac{1}{2}}(\mu), \end{equation} where $0<\mu<2$ and $K_\nu(x)$ is the modified Bessel function of the second kind. I did some simulation, it is monotone decreasing with respect to $n$, but so far have no idea how to prove it. Can you please help a bit? Thank you so much.

Answer 1

For $\mu>$ and $\nu>0$, we know that $K_\nu(\mu)>0$. We consider the ratio \begin{equation} \frac{p_n}{p_{n-1}}=\frac{\mu}{2n}\frac{K_{n-\tfrac{1}{2}}(\mu)}{K_{n-\tfrac{3}{2}}(\mu)} \end{equation} In theorem 1.2 of this paper, the following inequality is derived \begin{equation} \frac{K_{\nu}(t)}{K_{\nu-1}(t)}<\frac{\nu+\sqrt{\nu^2+t^2}}{t} \end{equation} for real $\nu$. Then \begin{equation} \frac{p_n}{p_{n-1}}<\frac{n-\tfrac{1}{2}+\sqrt{\left( n-\tfrac{1}{2} \right)^2+\mu^2}}{2n} \end{equation} It can be shown that, for $n>\mu^2+\tfrac{1}{4}$, rhs of the previous expression is an increasing function of $n$ bounded by 1 for $n\to \infty$. Thus $p_{n+1}<p_n$.

It can further be shown that rhs is lower than 1 for $n>\frac{\mu^2}{2}$. If $ \mu<2$ then the sequence decreases for $n>1$.