I recently came across the following question posed by a friend. Let $k\in\Bbb Z^+$ be a positive integer and consider the $(k+1)\times (k+1)$ matrix defined as
$$ A_k= \begin{pmatrix} 1 & \frac{1}{3} & \dots & \dots & \frac{1}{2k+1}\\ \frac{1}{3} & \dots & \dots & \frac{1}{2k+1} & \frac{1}{2k+3}\\ \frac{1}{5} & \dots &\frac{1}{2k+1}& \frac{1}{2k+3} &\frac{1}{2k+5}\\ \dots & \dots & \dots & \dots & \dots \\ \frac{1}{2k+1} & \dots & \dots &\dots & \frac{1}{4k+1} \end{pmatrix}. $$
For instance, $$A_1= \begin{pmatrix} 1 & \frac{1}{3}\\ \frac{1}{3} & \frac{1}{5}\\ \end{pmatrix}, $$
$$ A_2= \begin{pmatrix} 1 & \frac{1}{3} & \frac{1}{5}\\ \frac{1}{3} & \frac{1}{5} & \frac{1}{7}\\ \frac{1}{5} &\frac{1}{7}& \frac{1}{9}\\ \end{pmatrix}, $$
$$ A_3= \begin{pmatrix} 1 & \frac{1}{3} & \frac{1}{5} & \frac{1}{7}\\ \frac{1}{3} & \frac{1}{5} & \frac{1}{7} & \frac{1}{9}\\ \frac{1}{5} & \frac{1}{7}& \frac{1}{9} &\frac{1}{11}\\ \frac{1}{7} & \frac{1}{9}& \frac{1}{11}& \frac{1}{13} \end{pmatrix}. $$
My question is: Can we show all the matrices have non-zero determination for all $k\in\Bbb Z^+$?
A brute force computation shows that $\det(A_k)\neq0$ for $k=1,2,3,4$. Is there a simpler and smarter way for computing $\det(A_k)$ for a generic $k?$. In addition, looking at small values of $k$, it seems that $\det(A_k)>\det(A_{k+1})$. Is $$ \lim_{k\to\infty} \det(A_k)=0\,?$$
