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

Lesson 7: Alternating series test for convergence

# Worked example: alternating series

Example using the alternating series test to determine which values of a variable will make the series converge.

## Want to join the conversation?

• I understood how you took p/6 < 1, but can we not get the same result using the ratio test? That is, will the ratio test not be applicable here?
• Good question! The answer is almost, but not quite! Remember that the ratio test says that you should find (the absolute value of) the limit of the ratio of successive terms, and if THAT is < 1, then FOR SURE the series converges, but that if THAT is EQUAL to 1, then the test is INCONCLUSIVE. So in this case, it's true that the ratio test would say, "IF p/6 < 1, THEN it is certain that the series converges". BUT, the ratio test also says, "IF p/6 = 1, then I'm sorry but I can't say for sure what happens." Thus, the ratio test would not be able, alone, to choose between the first and second answer choices.
• I didn't understand what if the question didn't say find the POSITIVE values, what if he just wanted the values.

Does that imply that p could be negative?

and if it was negative then how would they know that it's decreasing so we can apply the alternating series test because bn must be decreasing
• Any values of p must not be negative for it to be an alternating series. By definition according to the Alternating Series Test, all of the b_sub_n terms (which are (p/6)^n in this case) must be greater than 0. The part about the positive values in the question was just thrown in as a hint.

If you DID consider p values that are negative, then (p/6)^n could be factored as (-1)^n * (-p)^n. Then the original problem would be:
``(-1)^n+1 * (-1)^n * (p/6)^n= (-1)^(2n+1) * (p/6)^n= (-1) * (p/6)^n    (since 2n+1 is always odd)= -(p/6)^n``

And that, is not an alternating series.
• Can`t we include 6?If P is 6, then doesn`t it converge to 0?
• With 𝑝 = 6, the sum keeps bouncing back and forth between 0 and 1, and doesn't converge to one specific value.
• Can someone help me verify if this is true?

When using the Alternating Series Test (AST), do I need to look at the absolute values of the terms and see if they converge to confirm that the series is absolutely convergent? If they don't converge, would the series be conditionally convergent?

Also, to prove if the absolute value of the series is convergent, what tests do I need to use? I've used the ratio and geometric tests, but are there others that are easier?

Thanks!
• No, you don't need to look at the absolute values. For instance the harmonic series (1 + 1/2 + 1/3 + ...) is divergent whereas the alternating series (1 - 1/2 + 1/3 - 1/4 + ..) Converges.

You can also use the integral test to prove convergence for the absolute values. The type of test used varies with the nature of the series.
• It says in the video that p can not be 6 because it wouldn't be monotonically decreasing. But isn't monotonically decreasing mean that that the proceeding term is less than or equal to the previous term? In order to use the alternating test, bn must be monotonically strictly decreasing, not just decreasing. 1 wouldn't work b/c it is a geometric series and |r|>= 1 is divergent.
• How to prove that the positive terms of an alternating series forms a Divergent series?
(1 vote)
• They don't necessarily form a divergent series. This is only true if the series is conditionally convergent, which means that the series converges if you add all the terms, but diverges if you add the absolute values of all the terms.
• Why can't p=0? (it can be greater than but not equal to).
When explaining the alternating series test in a previous video, Sal wrote B sub n can be greater than or equal to zero.
(1 vote)
• Have you read the question in the video?
(1 vote)
• At , Why does Sal write exponents just considering n? Does he neglect "plus one" we have next to the exponent "n"?