Sum of Elements in Inverse of Hilbert Matrix

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


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

Consider its inverse $H_n^{-1}$.

All the elements of $H_n^{-1}$ are integers.

The sum of all the elements $b_{i j}$ of $H_n^{-1}$ is:
 * $\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = n^2$

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 Sum of Elements in Inverse of Cauchy Matrix, the sum of all the elements of $H_n^{-1}$ is:
 * $\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$