Let $J\_\nu$ be a Bessel function of the first kind and let $\{\lambda\_{n, \nu}\}\_{n\ge 1}$ be a sequence of its zeroes. I claim that
$$
\inf\_{n\ge 1}\bigg|\sqrt{\lambda\_{n,\nu}} J\_{\nu+1}(\lambda\_{n,\nu})\bigg|>0.
$$
The reason I believe this is true is that:


(1) we have $|J\_{\nu+1}(x)|\lesssim x^{-1/2},\qquad x\rightarrow \infty$,


(2) there's an asymptotic formula
$$
\lambda\_{n,\nu}=\pi n+\frac{\pi(2\nu-1)}{4}+O(n^{-1}),
$$
which shows in particular that zeroes of $J\_{\nu+1}$ are separated from zeroes of $J\_\nu$, thus the only decay $J\_{\nu+1}(\lambda\_{n,\nu})$ is only due to (1) (i.e. oscillations should not be relevant).


I don't know how to make this argument rigorous. Any hints are appreciated!



Asymptotic behavior of a Bessel function on a sequence on zeros with a shifted parameter of type
You have, using the asymptotics for $\lambda\_{n,\nu}$,
\begin{align\*}
& \cos\left(\lambda\_{n,\nu} - \frac{2\nu+3}{4}\pi \right) \\
&= \cos\left(\lambda\_{n,\nu} - \frac{2\nu-1}{4}\pi+\pi \right) \\
&= \cos\left(\lambda\_{n,\nu} - \frac{2\nu-1}{4}\pi-n\pi \right) (-1)^{n-1}\\
&= \cos\left(O(n^{-1}) \right) (-1)^{n-1} \\
&= (-1)^{n-1} +O(n^{-1}).
\end{align\*}
Inserting this in the asymptotic formula for $J\_{\nu+1}(x)$, we obtain
$$
\sqrt{\lambda\_{\nu,n}}J\_{\nu+1}(\lambda\_{\nu,n})= (-1)^{n-1}\sqrt{\frac2\pi} +O(n^{-1})+O(\lambda\_{\nu,n}^{-1})=(-1)^{n-1}\sqrt{\frac2\pi} +O(n^{-1}).
$$
Which shows that
$$
\lim\_{n\to\infty} \lambda\_{\nu,n}J\_{\nu+1}^2(\lambda\_{\nu,n})=\frac2\pi.
$$
As $J\_{\nu}$ and $J\_{\nu+1}$ have distinct zeros, you have a proof.


