proof of comparison test

Assume |ak|bk for all k>n. Then we define




Obviously sktk for all k>n. Since by assumption (tk) is (tk) is bounded and so is (sk). Also (sk) is monotonic and therefore . Therefore i=0ai is absolutely convergent.

Now assume bkak for all k>n. If i=kbi is divergent then so is i=kai because otherwise we could apply the test we just proved and show that i=0bi is convergent, which is is not by assumption.

Title proof of comparison test
Canonical name ProofOfComparisonTest
Date of creation 2013-03-22 13:22:06
Last modified on 2013-03-22 13:22:06
Owner mathwizard (128)
Last modified by mathwizard (128)
Numerical id 4
Author mathwizard (128)
Entry type Proof
Classification msc 40A05