Inverse of Hilbert Matrix

Theorem
Let $H_n$ be the Hilbert matrix of order $n$:


 * $\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$

Then its inverse $H_n^{-1} = \sqbrk b_n$ can be specified as:


 * $\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - j}! \paren {n - i}! \paren {i + j - 1} } \end{bmatrix}$

Proof
From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix:


 * $\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$

where:
 * $x_i = i$
 * $y_j = j - 1$

From Inverse of Cauchy Matrix, the inverse of the square Cauchy matrix of order $n$ is:


 * $\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } } \end{bmatrix}$

Thus $H_n^{-1}$ can be specified as:


 * $\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} } {\ds \paren {i + j - 1} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} } } \end{bmatrix}$

First, from Product of Products:


 * $\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} = \prod_{k \mathop = 1}^n \paren {i + k - 1} \prod_{k \mathop = 1}^n \paren {j + k - 1}$

We address in turn the various factors of this expression for $b_{i j}$.

and similarly:
 * $\ds \prod_{k \mathop = 1}^n \paren {j + k - 1} = \frac {\paren {j + n - 1}!} {\paren {j - 1}!}$

Then:

and similarly:
 * $\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} = \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!$

Thus we can write: