Abel summation by parts: with $A_n = \sum_{k<n} a_k$ one has $\sum_{k<n} a_k b_k = A_n b_{n-1} - \sum_{k < n-1} A_{k+1}\,(b_{k+1} - b_k)$ for every $n \ge 1$

Let $(a_k)$ and $(b_k)$ be sequences of reals and let

$A_n \;:=\; \sum_{k<n} a_k \qquad (n \in \mathbb{N})$

be the partial sums of $\sum a_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion), so that $A_0 = 0$ and $a_k = A_{k+1} - A_k$ for every $k$. Then for every natural number $n \ge 1$

$\sum_{k<n} a_k b_k \;=\; A_n\, b_{n-1} \;-\; \sum_{k<n-1} A_{k+1}\,(b_{k+1} - b_k) .$

Both sides are finite sums in the sense of Finite sums and finite products, by recursion; at $n = 1$ the right-hand sum is empty and the identity reads $a_0 b_0 = A_1 b_0$.

The hypothesis $n \ge 1$ is what makes the statement legitimate, not merely convenient: the index $n-1$ occurs on the right, and $n-1$ is a natural number exactly when $n \ge 1$. At $n = 0$ there is nothing to state, both the left-hand side and $A_0$ being $0$.

• What the identity is for. It converts a series $\sum a_k b_k$, about which nothing is assumed, into a boundary term $A_n b_{n-1}$ and a series $\sum A_{k+1}(b_{k+1} - b_k)$ whose terms carry the differences of $(b_k)$. If $(A_n)$ is bounded and $(b_k)$ is monotone, those differences have one sign and telescope, which is exactly the situation of Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$, then $\sum a_k b_k$ converges. The transformation is the discrete analogue of integration by parts, and the boundary term is the analogue of the boundary term there.

• The block form needs no separate proof. For $1 \le M \le n$, subtracting the identity at $M$ from the identity at $n$ gives $\sum_{k=M}^{n-1} a_k b_k = A_n b_{n-1} - A_M b_{M-1} - \sum_{k=M-1}^{n-2} A_{k+1}(b_{k+1}-b_k)$, using only splitting of finite sums (Laws of finite sums and finite products). Nothing on this page needs that form, so it is recorded here rather than stated as a result.

• Two conventions are doing work. $A_n$ sums the $n$ terms $a_0, \dots, a_{n-1}$, so $A_0 = 0$ and $a_k = A_{k+1} - A_k$ with no shift (Series, partial sums, convergence and the sum, divergence, and the tail series); and the empty sum is $0$ (Finite sums and finite products, by recursion), which is what makes $n = 1$ a genuine instance of the identity rather than a case to be excluded.