While working on a problem, I managed to reduce it to showing that the matrix $A = [a_{ij}]$, where $2a_{ij} = \frac{1}{i+j-1} - \frac{(-1)^{i+j-1}}{i+j-1}$, has non-zero determinant, for whatever size you choose. Here are a few cases: $$\det\begin{bmatrix} 1 & 0 & \frac{1}{3} \\ 0 & \frac{1}{3} & 0 \\ \frac{1}{3} & 0 & \frac{1}{5} \end{bmatrix} = \frac{4}{135};$$
$$\det\begin{bmatrix} 1 & 0 & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & 0 & \frac{1}{5}\\ \frac{1}{3} & 0 & \frac{1}{5} & 0 \\ 0 & \frac{1}{5} & 0 & \frac{1}{7} \end{bmatrix} = \frac{16}{23625}$$
I worked on a few cases using online calculators, but explicit determinant calculation in arbitrary sizes is very cumbersome. It seems to tend to $0$, but to always be positive. In fact, I suspect this is a positive-definite matrix, but, again, couldn't quite prove it - and I recall having seen it in a computational setting before, maybe numerical integration, though I'm not sure...
Anyhow, is there any technique I can use to show this has non-zero determinant for any size I pick?
Thanks in advance!
Let $C:=(\frac1{i+j-1})_{1\le i,j\le n}$. This is a symmetric matrix known as a Cauchy matrix. By Sylvester's criterion, it is positive-definite.
Let $D$ be the $n\times n$ diagonal matrix whose $i$-th diagonal element is $(-1)^{i-1}$. We can easily check by matrix multiplication that $A=(DCD+C)/2$.
We observe that $DCD=(D\sqrt C)(D\sqrt C)^T$ where $\det(D\sqrt C)\neq0$, so $DCD$ is also a positive-definite matrix. Hence $A$ is the sum of two positive-definite matrices, and therefore also positive-definite. In particular $\det(A)>0$.