<s>[INST] How to prove the determinant of a Hilbert-like matrix with parameter is non-zero Consider some positive non-integer $\beta$ and a non-negative integer $p$. Does anyone have any idea how to show that the determinant of the following matrix is non-zero?
$$
\begin{pmatrix}
\frac{1}{\beta + 1} & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{p+1}\\
\frac{1}{\beta + 2} & \frac{1}{3} & \frac{1}{4} & \dots & \frac{1}{p+2}\\
\frac{1}{\beta + 3} & \frac{1}{4} & \frac{1}{5} & \dots & \frac{1}{p+3}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{\beta + p + 1} & \frac{1}{p+2} & \frac{1}{p+3} & \dots & \frac{1}{2p+1}
\end{pmatrix}.
$$
 [/INST] Rows linearly dependent means for some $c_1$, $\ldots$, $c_{p+1}$ the non-zero rational function 
$\sum_{k=1}^{p+1}  \frac{c_k}{x+k}$ has $p+1$ roots $\beta$, $1$, $2$, $\ldots$, $p$, not possible, since its numerator has degree at most $p$.
</s>