It can easy be shown by induction that the determinant of the $(N-1)\times (N-1)$ matrix $$\begin{pmatrix} 2 & -1 & & \\ -1 & 2 & \ddots & \\ & \ddots & \ddots & -1\\ & & -1 & 2 \end{pmatrix}$$ equals $N$. (Up to a factor) this matrix corresponds to the representing matrix of the discrete Laplacian on an interval with Dirichlet boundary conditions, on an equidistant partition.

Now it the partition is not equidistant, say it is $0 = \tau_0 < \tau_1 < \dots < \tau_N$ with increments $\Delta_j := \tau_j - \tau_{j-1}$, then (up to a factor) we end up with the matrix $$\begin{pmatrix} \Delta_1 + \Delta_2 & -\Delta_2 & & \\ -\Delta_2 & \Delta_2 + \Delta_3 & \ddots & \\ & \ddots & \ddots & -\Delta_{N-1}\\ & & -\Delta_{N-1} & \Delta_{N-1}+\Delta_N \end{pmatrix}.$$ Is the determinant of this matrix still explicitly computable? If not, can we at least compute some (suitably normalized) limit of the determinant as the mesh of the partition goes to zero?

\edit: user35593 answered in the comments that the determinant is equal to $$ \prod_{j=1}^N {\Delta_j}\sum_{i=1}^N\frac{1}{\Delta_i},$$ which is indeed easy to check by induction. Thank you very much!