Abel’s lemma

Theorem 1 Let${\{a_i\}}_{i=0}^N$ and ${\{b_i\}}_{i=0}^N$ be sequences ofreal (or complex) numbers with $N\ge 0$.For $n=0,\mathrm{\dots },N$, let $A_n$ be the partial sum$A_n={\sum }_{i=0}^n{a}_i$.Then

$$\sum _{i=0}^N{a}_i{b}_i=\sum _{i=0}^{N-1}{A}_i(b_i-b_{i+1})+A_N{b}_N.$$

In the trivial case, when $N=0$, then sum on the right hand sideshould be interpreted as identically zero. In other words,if the upper limit is below the lower limit, there is no summation.

An inductive proof can be found here (http://planetmath.org/ProofOfAbelsLemmaByInduction).The result can be found in [1] (Exercise 3.3.5).

If the sequences are indexed from $M$ to $N$, we have the followingvariant:

CorollaryLet ${\{a_i\}}_{i=M}^N$ and ${\{b_i\}}_{i=M}^N$ be sequences ofreal (or complex) numbers with $0\le M\le N$.For $n=M,\mathrm{\dots },N$, let $A_n$ be the partial sum$A_n={\sum }_{i=M}^n{a}_i$.Then

$$\sum _{i=M}^N{a}_i{b}_i=\sum _{i=M}^{N-1}{A}_i(b_i-b_{i+1})+A_N{b}_N.$$

Proof. By defining$a_0=\mathrm{\dots }=a_{M-1}=b_0=\mathrm{\dots }=b_{M-1}=0$, we can apply Theorem 1to the sequences ${\{a_i\}}_{i=0}^N$ and ${\{b_i\}}_{i=0}^N$.$\mathrm{\square }$