1. Limit Comparison Test Clarification

Okay so, the limit as n goes to infinity of a_n/b_n = L >0; If sum of b_n converges then the sum of a_n also converges and if sum of b_n diverges then sum of a_n diverges.

Zero-Case: If the limit as n goes to infinity of a_n/b_n = 0; if sum of b_n converges then the sum of a_n also converges. Is it true that: the limit as n goes to infinity of a_n/b_n = 0; if the sum of b_n diverges then the sum of a_n also diverges? Is it also true that: the limit as n goes to infinity of b_n/a_n = 0; if the sum of b_n converges (or diverges) then the sum of a_n converges (or diverges)?

Infinity-Case: if the limit as n goes to infinity of a_n/b_n = infinity; if the sum of b_n converges then the sum of a_n converges. Is this true: if the sum of b_n diverges then the sum of a_n also diverges? Is it also true that: the limit as n goes to infinity of b_n/a_n = infinity; if the sum of b_n converges (or diverges) then the sum of a_n converges (or diverges)?

Can someone confirm or correct me on these statements, please?

2. Re: Limit Comparison Test Clarification

Hey Preston019.

You mention two sequences a_n and b_n where a_n/b_n = L > 0 for all n. So a_n = L*b_n where L is fixed. Now you want to show if a_n converges then so does b_n and if it diverges then so does the other.

The easiest way IMO is to just put sigma signs and sum the terms to infinity and then take the L out of the series and then you have an expression where A = L*B where A is the series for a_n, B is the series for b_N and L is a constant > 0.

The rigorous way of answering this involves the delta-epsilon stuff but there are some basic identities that show if X is a convergent series (in sigma form) then cX is also convergent as well. The same holds for cX being divergent if X is divergent (but c must be a constant that is non-zero).

Again I don't know what expectations you are under but you show the above two (i.e. stuff related to cX where X is convergent/divergent and c is a non-zero constant) then that's your proof.