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A telescoping sum cancels intermediate sequence terms, leaving boundary terms. For a finite sequence of real numbers $a_0,\dots ,a_n$, the displayed identity follows because every $a_1,\dots ,a_{n-1}$ occurs once with each sign. Applying it to an infinite series also requires checking the limit of the boundary terms; finite cancellation alone does not prove that the series is a convergent series.^[ref]

For $n\ge 1$, expanding the finite sum givesNo monotone sequence assumption is required. Consider the original sequence $4,9,2,7,5$. Its successive finite differences are $5,-7,5,-2$, soThe large intermediate rise and fall disappear from the net change. This is useful whenever one knows successive changes rather than every absolute value: adding the changes reconstructs the final value from the initial one. It also checks a cumulative-update calculation without requiring all increments to be positive.

A finite difference $\Delta a_{k-1}=a_k-a_{k-1}$ measures one discrete change. The identity ${\sum }_{k=1}^n\Delta a_{k-1}=a_n-a_0$ is the discrete counterpart of the fundamental theorem of calculus: accumulation of changes returns the endpoint difference. The related summation by parts transfers a finite difference from one factor of a product to another, with explicit boundary terms.

Adding a total derivative $dg(q(t))/dt$ to a Lagrangian changes its action only by $g(q(T))-g(q(0))$. Under fixed-endpoint variations, this boundary term is constant, so the Euler-Lagrange equations and their minimizing paths are unchanged. Componentwise, $d[\partial (dg/dt)/\partial {\dot{q}}_i]/dt={\sum }_j{g}_{ij}{\dot{q}}_j=\partial (dg/dt)/\partial q_i$ by equality of mixed partial derivatives. Endpoint freedom would require reconsidering the resulting natural boundary conditions for a free endpoint.