Abel's inequality
# Abel's inequality

In mathematics, Abel's inequality, named after Niels Henrik Abel, supplies a simple bound on the absolute value of the inner product of two vectors in an important special case.

## Mathematical description

Let ${a_1,$ $a_2,...}$ be a sequence of real numbers that is either nonincreasing or nondecreasing, and let ${b_1,$ $b_2,...}$ be a sequence of real or complex numbers. If ${a_n}$ is nondecreasing, it holds that

    $\left|\sum _{k=1}^{n}a_{k}b_{k}\right|\leq \operatorname {max} _{k=1,\dots ,n}|B_{k}|(|a_{n}|+a_{n}-a_{1}),$

and if ${a_n}$ is nonincreasing, it holds that

    $\left|\sum _{k=1}^{n}a_{k}b_{k}\right|\leq \operatorname {max} _{k=1,\dots ,n}|B_{k}|(|a_{n}|-a_{n}+a_{1}),$

where

    $B_{k}=b_{1}+\cdots +b_{k}.$

In particular, if the sequence ${a_n}$ is nonincreasing and nonnegative, it follows that

    $\left|\sum _{k=1}^{n}a_{k}b_{k}\right|\leq \operatorname {max} _{k=1,\dots ,n}|B_{k}|a_{1},$

## Relation to Abel's transformation

Abel's inequality follows easily from Abel's transformation, which is the discrete version of integration by parts: If ${a_1,$ $a_2,$ ...} and ${b_1,$ $b_2,$ ...} are sequences of real or complex numbers, it holds that

    $\sum _{k=1}^{n}a_{k}b_{k}=a_{n}B_{n}-\sum _{k=1}^{n-1}B_{k}(a_{k+1}-a_{k}).$
