# 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.

Let {a1, a2,...} be a sequence of real numbers that is either nonincreasing or nondecreasing, and let {b1, b2,...} be a sequence of real or complex numbers. If {an} is nondecreasing, it holds that

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

and if {an} is nonincreasing, it holds that

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

where

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

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

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

Abel's inequality follows easily from Abel's transformation, which is the discrete version of integration by parts: If {a1, a2,...} and {b1, b2,...} are sequences of real or complex numbers, it holds that

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