I would like to prove the following version of Abel summation $$\sum_{k=1}^nu_kv_k = u_{n+1}\sum_{k=1}^nv_k - \sum_{k=1}^n\Delta u_k\left[ \sum_{p=1}^kv_p\right]$$ where $\Delta u_k = u_{k+1} - u_k.$ I can see how to unravel this sum by hand and show this equality, but I'm trying to construct a proof by discrete calculus operator identities and the fundamental theorem of calculus for discrete functions: $\sum_{k=1}^nu_k = (\Delta^{-1}u)_{n+1}-(\Delta^{-1}u)_{1}$
My attempt:
We have the product formula: $$\Delta^{-1}(u\Delta w) = uw - \Delta^{-1}(Ew\Delta u)$$ with $Eu_k = u_{k+1}$ Set $v= \Delta w.$ Then $$\Delta^{-1}(uv) = u\Delta^{-1}v - \Delta^{-1}(\Delta u E \Delta^{-1}v)$$ or $$\sum_{k=1}^nu_kv_k = (u_{n+1}- u_1)\sum_{k=1}^n v_k - \sum_{k=1}^n \Delta u_k (\Delta^{-1}v)_{k+1}$$ If correct, this seems very close to what I'd like to show but I don't see how to complete the argument. Is there a way to elegantly obtain this formula via discrete calculus operators?
We can show the formula using summation by parts \begin{align*} \sum_{k=1}^n a_k\Delta b_k =\big[a_k b_k\big]_{k=1}^{n+1}-\sum_{k=1}^n b_{k+1}\Delta a_k\tag{1} \end{align*} In order to apply (1) we set \begin{align*} V_k:=\begin{cases} v_1+v_2+\cdots+v_{k-1}&\qquad k\geq 2\\ 0&\qquad k=1 \end{cases}\tag{2} \end{align*} so that we can use $\color{blue}{v_k=\Delta V_k, k\geq 1}$.