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:

$\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = n^2$

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:

$x_i = i$
$y_j = j - 1$

Then:

 $\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}$ $=$ $\ds \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$ Sum of Elements in Inverse of Cauchy Matrix $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n k + \sum_{k \mathop = 1}^n \paren {k - 1}$ $\ds$ $=$ $\ds 2 \sum_{k \mathop = 1}^n k - n$ $\ds$ $=$ $\ds n \paren {n + 1} - n$ Closed Form for Triangular Numbers $\ds$ $=$ $\ds n^2$

$\blacksquare$