Sum of Elements in Inverse of Hilbert Matrix

Theorem

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

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:

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

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

where:

$x_i = i$

$y_j = j - 1$

Then:

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

$=$

$\ds \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$

$\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 $

$=$

$\ds n \paren {n + 1} - n$

$\ds $

$=$

$\ds n^2$

$\blacksquare$