TITLE: How prove this determinant is $0?$
QUESTION [5 upvotes]: find the value
$$A_{n}=\begin{vmatrix}
1-\dfrac{1}{(n+1)^2}&\dfrac{1}{2}&\dfrac{1}{3}&\cdots&\dfrac{1}{n+1}\\
\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}&\cdots&\dfrac{1}{n+2}\\
\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}&\cdots&\dfrac{1}{n+3}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\dfrac{1}{n+1}&\dfrac{1}{n+2}&\dfrac{1}{n+3}&\cdots&\dfrac{1}{2n+1}
\end{vmatrix}$$
show that $\det(A_{n})=0$
I have prove
$$\det(A_{1})=\begin{vmatrix}
\dfrac{3}{4}&\dfrac{1}{2}\\
\dfrac{1}{2}&\dfrac{1}{3}
\end{vmatrix}=0$$
and 
$$\det(A_{2})=\begin{vmatrix}
\dfrac{8}{9}&\dfrac{1}{2}&\dfrac{1}{3}\\
\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}\\
\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}
\end{vmatrix}=0$$
$$\det(A_{3})=\begin{vmatrix}
\dfrac{15}{16}&\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}\\
\dfrac{1}{2}&\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}\\
\dfrac{1}{3}&\dfrac{1}{4}&\dfrac{1}{5}&\dfrac{1}{6}\\
\dfrac{1}{4}&\dfrac{1}{5}&\dfrac{1}{6}&\dfrac{!}{7}
\end{vmatrix}=0$$

REPLY [4 votes]: Shifting the index, we have
$$A_n = H_n -(1/n^2) e_1 e_1^T$$
where $H_n$ is the Hilbert matrix and $e_1$ is the column vector $[1,0,0,\dots,0]^T$.
By the matrix determinant lemma,
$$
\det(A_n)=(1-(1/n^2)e_1^T H_n^{-1} e_1) \det(H_n)
=(1-(1/n^2) (H_n^{-1})_{11}) \det(H_n) = 0
$$
because the $1,1$ entry of $H_n^{-1}$ is $n^2$.