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Comparison Test Part II I. Quiz II. Homework III. Proof of the Comparison Test Let an be sequence of positive numbers such that 0 < bn < an and such that the sequence sum an = L then if Bn represents the nth partial sum of bn and An is the nth partial sum of an then 0 < Bn < An < L so that Bn is a bounded sequence. Bn is monotone since the terms are all positive, hence Bn converges. Now let an be a sequence of positive numbers such that 0 < an < bn such that sum an diverges. Then if bn converges this would contradict the first part of the Comparison test with the roles of a and b switched. Hence bn diverges.
IV. Proof of the Limit Comparison Test Suppose that sum bn diverges and that lim an /bn = L > 0, then for large n, an >bn (L/2) , but if sum bn diverges so does sum bn (L/2) Now by the direct comparison test, sum an diverges V. Notes on the Limit Comparison Test If bn converges and lim an /bn = 0 then an is forced to be very small compared to bn for large n and hence sum an also diverges. Also if bn diverges and lim an /bn = infinity then an is forced to be very large hence sum an also diverges. |