# computational mathematics – Determinant of certain symmetric matrices

I recently came across the following question posed by a friend. Let $$kinBbb Z^+$$ 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 $$kinBbb Z^+$$?

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
$$lim_{ktoinfty} det(A_k)=0,?$$