Let $t>0$, and for $k=0,1,2,...,$ let $ b_k=k^t, a_k=(b_k+b_{k+1})/2$, and set $$A(n,t)=\prod_{k=0}^{n-1}\frac{b_n-a_k}{b_n-b_k}.$$ How does one prove (or disprove) that for any fixed $t>0$, $A(n,t)$ is decreasing and approach $0$ as $n \rightarrow \infty$ ? The result is true for $t=1$ since $A(n,1)=\frac{{2n-1 \choose n-1}}{2^{2n-1}}$, and numerical computations suggests it holds for all $t>0$. It seems to require different convexity for $t \ge 1 $ and $t \le 1$.
Motivation: Since $0<A(n,t)<1$, we know all the polynomials $$u_{n,t}(x)=\prod_{k=0}^{n-1}(x+a_k)-A(n,t) \prod_{k=0}^{n-1}(x+b_k),$$ has only real and simple roots by arguments in this previous MO (for the case $t=1$).
"Interlacing property" of certain polynomials
The same argument will imply that the zeros of $u_{n,t}(x)$ interlace if $A(n,t)$ is decreasing, and the $k$th largest root $\lambda_{n,k}$ will converge to $-a_{k-1}$ if we know $A(n,t) \rightarrow 0$. Note $u_{n,t}(-b_n)=0$ as in the case $t=1$.
A more general question is if we only know $0=b_0<a_0<b_1<a_1<b_2<a_2<... \rightarrow \infty$ are two interlacing sequence of real numbers, in which case $u_{n,t}(x)$ still have only real and simple roots. Are there general conditions on $a_n,b_n$ to ensure $A(n,t)$ will be decreasing and approach zero?
