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### Course: AP®︎/College Calculus BC>Unit 10

Lesson 6: Comparison tests for convergence

# Worked example: limit comparison test

To use the limit comparison test for a series S₁, we need to find another series S₂ that is similar in structure (so the infinite limit of S₁/S₂ is finite) and whose convergence is already determined. See a worked example of using the test in this video.

## Want to join the conversation?

• Does it matter which we set as "a" and "b" during the limit comparison test?
• Technically, it doesn't matter, because the only requirements for the limit is that it is finite and positive (0 < c < infinity). Because the exact value is not required for this test, it does not matter which is 'a' and which is 'b' for this type of problem.
• Why is it that, even though (2/3)^n is smaller than (2^n)/(3^n)-1, Sal still used it for the comparison? I would have understood if it diverged, because a>b in that case, but since it is converging, isn't b>a supposedly?
• That concept is applied to the direct comparison test. For the limit comparison test, it does not matter if one series is greater than or less than the other series; as long as the limit of their ratios approach a positive finite value, then they are either both convergent or both divergent.
• I really do feel like these videos are fabulous for anyone in Freshman Calculus wanting to further develop their basic intuition a little bit.
• Don't all the functions meet the criteria specified? I realize that the chosen one corresponds the behavior better, but that's rather abstract. But surely the other functions are also zero or greater?
• Is there a way to find out what the original S converges to?
• The behaviour of the infinite terms of a series is enough to know whether a series converges or diverges ??
(1 vote)
• No, not in general.

But ∑(2∕3)^𝑛 is a geometric series with a common ratio of 2∕3,
which is less than 1, and thereby the series converges.

By the comparison test we can then conclude that
∑2^𝑛∕(3^𝑛 − 1) also converges.
• At to , Sal says that series a sub(n) and b sub(n) are greater than or equal to zero. Just to clarify b sub(n) cannot be equal to zero it is only greater than zero since it's the denominator.
(1 vote)
• There's nothing saying that 𝑏(𝑛) can not be equal to 0 for some value/s of 𝑛, because we only care about the limit of 𝑎(𝑛)∕𝑏(𝑛) as 𝑛 approaches infinity, and as the video clearly shows, this limit can exist even if 𝑏(𝑛) tends to 0.