Limit comparison test

A few people have asked about the limit comparison test during office hours. It’s stated on page 564 of your text, but we can reword it in a way that might make it easier to understand.

First, imagineĀ that instead of assuming that \lim_{n \to \infty} \frac{a_n}{b_n} = L, we assumed that Ā \frac{a_n}{b_n} = L for all values of n, not just in the limit.

If 0 < L < \infty, we could write a_n = L b_n and b_n = \frac{1}{L}. Either way, a_n is a multiple of b_n and vice versa. Multiplying a series by a constant (non-zero, non-infinite) doesn’t change whether it converges or diverges. Therefore, either both series \sum a_n and \sum b_n converge or they both diverge….both series have to behave the same way.

If L isĀ very very large, then b_n = \frac{1}{L} a_n, that is, b_n is a tiny multiple of a_n. Therefore, if you know that \sum a_n converges, then \sum b_n must also converge.

If L is very very small, then a_n = L b_n, that is, a_n is a tiny multiple of b_n. Therefore, if you know that \sum b_n converges, then \sum a_n must also converge.

None of what I’ve said so far should sound too crazy. What’s interesting about the limit comparison test is that you don’t have to assume that \frac{a_n}{b_n} = L for all values of n, you just have to assume thatĀ \frac{a_n}{b_n} is close to L only for large values of n. IfĀ 0 < L < \infty, what this says is that a_n is eventually almost a multiple of b_n and vice versa.Ā Therefore, either both series \sum a_n and \sum b_n converge or they both diverge…both series have to behave the same way.

Additionally, if you let L actually approachĀ \infty, thenĀ b_n is eventually almost a tiny multiple of a_n.Ā Therefore, if you know that \sum a_n converges, then \sum b_n must also converge.

Finally, if you let L actually approach 0 , then a_n is eventually almost a tiny multiple of b_n.Ā Therefore, if you know that \sum b_n converges, then \sum a_n must also converge.

 

tl;dr: If the individual termsĀ of one series are eventually almost equal to a multiple of the individual terms of another series, then both series have to behave the same way (both converge or both diverge).

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