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 $H_n^{-1} = \left[{b}\right]_n$ can be specified as:


 * $\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\displaystyle \prod_{k \mathop = 1}^n \left({i + k - 1}\right) \left({j + k - 1}\right)} {\displaystyle \left({i + j - 1}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \left({i - k}\right)}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({j - k}\right)}\right)} \end{bmatrix}$

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:


 * $\begin{bmatrix} b_{ij} \end{bmatrix} = \begin{bmatrix} \dfrac {\displaystyle \prod_{k \mathop = 1}^n \left({x_j + y_k}\right) \left({x_k + y_i}\right)} {\displaystyle \left({x_j + y_i}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)}\right) \left({\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \left({y_i - x_k}\right)}\right)} \end{bmatrix}$

Substituting $r$ for $x_r$ and $r - 1$ for $y_r$ (according to index) throughout yields the result, after rearrangement.