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