We attempt to prove the following conjecture:

For the infinite symmetric matrix

\[C_{ij} = ij (\frac{1}{i + j + 1} + \frac{1}{i + j - 1})\]

the inverse obeys:

\[\sum_{i = 1, j = 1}^{\infty, \infty} C_{ij}^{-1} = \frac{\pi}{4}\]

For reference, the first few entries of this matrix look like this:

\[\begin{pmatrix} 1(1/1 + 1/3) & 2 (1/2 + 1/4) & 3 (1/3 + 1/5) & \ldots \\ 2(1/2 + 1/4) & 4 (1/3 + 1/5) & 6 (1/4 + 1/6) & \ldots \\ 3(1/3 + 1/5) & 6 (1/4 + 1/6) & 9 (1/5 + 1/7) & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}\]

First we explicitly construct the inverse. To do this, break the matrix into a bunch of smaller matrices that are easier to invert. First, we can remove the factors of \(ij\) by multiplying a matrix \(D\) on each side by \(N = \text{diag}(1,2,3,4, \ldots)\):

\[C = N \begin{pmatrix} 1/1 + 1/3 & 1/2 + 1/4 & 1/3 + 1/5 & \ldots \\ 1/2 + 1/4 & 1/3 + 1/5 & 1/4 + 1/6 & \ldots \\ 1/3 + 1/5 & 1/4 + 1/6 & 1/5 + 1/7 & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix} N = NDN\]

\(D\) clearly consists of the sum of a matrix \(H\) added to itself, shifted over two rows by another matrix which we’ll call \(T\). So:

\[D = \begin{pmatrix} 1 & 0 & 1 & 0 & 0 & \ldots \\ 0 & 1 & 0 & 1 & 0 & \ldots \\ 0 & 0 & 1 & 0 & 1 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} \begin{pmatrix} 1/1 & 1/2 & 1/3 & \ldots \\ 1/2 & 1/3 & 1/4 & \ldots \\ 1/3 & 1/4 & 1/5 & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix} = TH\]

So far, \(N\) and \(T\) are easy to invert. The inverse of \(N\) is just \(\text{diag}(1/1, 1/2, 1/3, \ldots)\). The inverse of \(T\) is seen by considering it as the matrix form of the operator \((1 + x^2)\) (consider its action on the vector \((1, x, x^2, x^3, \ldots) \mapsto (1 + x^2, x + x^3, x^2 + x^4, \ldots)\). The inverse is the matrix form of the operator \(\frac{1}{1 + x^2} = \sum (-1)^{n} x^{2n} = 1 - x^2 + x^4 - x^6 + \ldots\), which is:

\[T^{-1} = \begin{pmatrix} 1 & 0 & -1 & 0 & 1 & \ldots \\ 0 & 1 & 0 & -1 & 0 & \ldots \\ 0 & 0 & 1 & 0 & -1 & \ldots \\ 0 & 0 & 0 & 1 & 0 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}\]

I was quite stumped on inverting \(H\) until I … googled it. Turns out it’s a well-known object called the Hilbert matrix, and there’s a ton written about it. And the inverse, at least of the finite \(n \times n\) version is some ridiculous expression:

\[H^{-1}_{ij} = (-1)^{i+j} (i + j - 1) { n + i - 1 \choose n - j } {n + j -1 \choose n -i } {i + j -2 \choose i -1 }^2\]

Which I have zero hope of using in an actual equation. I can’t find an explicit form for the inverse of the infinite version, but it’s probably okay to assume it has the same properties:

• alternating signs
• giant, basically-infinite integers everywhere
• a determine that gets really tiny

Although this text suggests that it doesn’t have a formal inverse … except also it’s complicated, as infinite matrices don’t even have the same rules; their multiplication isn’t necessarily associative (!). Apparently it has an inverse if restricted to \(\mathcal{l}^2\).

So anyway we have \(C = N TH N\) and thus

\[C^{-1} = N^{-1} H^{-1} T^{-1} N^{-1}\]

Assuming I got all of this right.

If those exist and if the rules of infinite matrices are the same as finite ones (no idea; presumably not quite for the same reason that you can do crazy things with infinite sums). Finally, the sum of all the entries in a matrix is a weird thing to compute, but it is given by \(\sum_{ij} Q_{ij} = \b{1}^T Q \b{1}\), where \(\b{1}\) is the vector \((1, 1, 1, 1, \ldots)\). So I guess what we want is \(\sum_{ij} C^{-1}_{ij} = \b{1}^TN^{-1} T^{-1} H^{-1} N^{-1} \b{1}\). But I am daunted at the task of handling the infinite inverse of \(H\).

One thing I did notice, though, is that \(\pi\) comes up a lot in discussions of the spectrum of \(H\). For instance: \(0 \leq x^T H x \leq \pi \| x \|^2\). So it’s not inconceivable that it’s related.