Dirichlet’s convergence test

Theorem. Let {an} and {bn} be sequences of real numbers such that {i=0nai} is boundedPlanetmathPlanetmathPlanetmath and {bn} decreases with 0 as limit. Then n=0anbn converges.

Proof. Let An:=i=0nan and let M be an upper bound for {|An|}. By Abel’s lemma,

i=mnaibi = i=0naibi-i=0m-1aibi
= i=0n-1Ai(bi-bi+1)-i=0m-2Ai(bi-bi+1)+Anbn-Am-1bm-1
= i=m-1n-1Ai(bi-bi+1)+Anbn-Am-1bm-1
|i=mnaibi| i=m-1n-1|Ai(bi-bi+1)|+|Anbn|+|Am-1bm-1|

Since {bn} converges to 0, there is an N(ϵ) such that both i=m-1n-1(bi-bi+1)<ϵ3M and bi<ϵ3M for m,n>N(ϵ). Then, for m,n>N(ϵ), |i=mnaibi|<ϵ and anbn converges.

Title Dirichlet’s convergence test
Canonical name DirichletsConvergenceTest
Date of creation 2013-03-22 13:19:53
Last modified on 2013-03-22 13:19:53
Owner lieven (1075)
Last modified by lieven (1075)
Numerical id 5
Author lieven (1075)
Entry type Theorem
Classification msc 40A05