proof of limit comparison test

The main theorem we will use is the comparison testMathworldPlanetmath, which basically states that if an>0, bn>0 and there is an N such that for all n>N, an<bn , then if i=1bn convergesPlanetmathPlanetmath so will i=1an.

Suppose limnanbn=L where L can be a non negative real number or +.

By definition, for L finite, this means that for every ϵ>0 there is a natural numberMathworldPlanetmath nϵ such that for all n>nϵ, anbn-L<ϵ

To make matters more concrete choose ϵ=L2 and assume L0 and finite.

0<an<3L2bn, for all n>nL2.

If i=1bn converges, so will i=13L2bn and thus by the comparison test, i=1an will also be convergentMathworldPlanetmathPlanetmath.

For the reverse result, consider limnbnan=1L, since if L is finite so will 1L, applying the previous result we can say that if i=1an converges so will i=1bn

Consider the case L=0, clearly L=0+ since both an and bn are positive, this means that for all ϵ>0 there exists nϵ such that for all n>nϵ, 0<an<ϵbn.

Considering ϵ=1 we get the exact formulation of the comparison test, so if i=1bn converges so will i=1an.

For the case L=+ just apply the result to limnbnan=0 to conclude that if i=1an converges so will i=1bn

Title proof of limit comparison test
Canonical name ProofOfLimitComparisonTest
Date of creation 2013-03-22 15:35:54
Last modified on 2013-03-22 15:35:54
Owner cvalente (11260)
Last modified by cvalente (11260)
Numerical id 4
Author cvalente (11260)
Entry type Proof
Classification msc 40-00