# Abel’s lemma

Let $\{a_{i}\}_{i=0}^{N}$ and $\{b_{i}\}_{i=0}^{N}$ be sequences of real (or complex) numbers with $N\geq 0$. For $n=0,\ldots,N$, 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 side should 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 following variant:

Corollary Let $\{a_{i}\}_{i=M}^{N}$ and $\{b_{i}\}_{i=M}^{N}$ be sequences of real (or complex) numbers with $0\leq M\leq N$. For $n=M,\ldots,N$, 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 $a_{0}=\ldots=a_{M-1}=b_{0}=\ldots=b_{M-1}=0$, we can apply Theorem 1 to the sequences $\{a_{i}\}_{i=0}^{N}$ and $\{b_{i}\}_{i=0}^{N}$. $\Box$

## References

Title Abel’s lemma AbelsLemma 2013-03-22 13:19:49 2013-03-22 13:19:49 mathcam (2727) mathcam (2727) 14 mathcam (2727) Theorem msc 40A05 summation by parts Abel’s partial summation Abel’s identity Abel’s transformation PartialSummation