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Calculus, all content (2017 edition)
Course: Calculus, all content (2017 edition) > Unit 7
Lesson 10: Ratio & alternating series testsWorked example: alternating series
Example using the alternating series test to determine which values of a variable will make the series converge.
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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?(6 votes)
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.(8 votes)
- 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(4 votes)
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.(5 votes)
Cant we include 6?t it converge to 0?
If P is 6, then doesn(2 votes)
With 𝑝 = 6, the sum keeps bouncing back and forth between 0 and 1, and doesn't converge to one specific value.(3 votes)
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!(2 votes)
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.(2 votes)
- 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.(2 votes)
- 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)
- 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.(1 vote)
- At0:28, Why does Sal write exponents just considering n? Does he neglect "plus one" we have next to the exponent "n"?
In other words, shouldn't the series start with exponent (p/6)ˆ2, (p/6)ˆ3...(1 vote) 
Does choosing between (-1)^(n+1) or (-1)^(n) matter when testing convergence or divergence?(0 votes)- No. Multiplying each term by -1 will not affect convergence or divergence.(1 vote)
0:28Video transcript
- [Instructor] What
are all positive values of p such that the series converges? So let's see. We have the sum from n equals one to infinity of negative one to the n plus one, times p over six to the n. So there's a couple of things
that might jump out at you. This negative one to the n plus one, as n goes from one to two to three, this is just going to
alternate between positive one, negative one, positive one, negative one. So we're gonna have alternating signs. That might be a little bit
of a clue of what's going on, and actually, let's just write it out. This is going to be,
see when n equals one, this is going to be to the second power, so it's gonna be positive one. So it's gonna be p over
six, and when n is two, this is gonna be to the third power, so it's gonna be minus p over six squared, then plus p over six to the third power. And I could even write to the
first power right over here. Then minus p over six to the fourth power, and we're gonna just
keep going plus, minus, on and on and on and on forever. This is clearly, this is a
classic alternating series right over here, and we could actually apply our alternating series test. And our alternating series test tells us that if this part of our expression, the part that is not alternating in sign, I guess you could say, if this part of the expression is
monotonically decreasing, monotonically decreasing, which is just a fancy way of saying that each successive term is less
than the term before it, and if we also know that the limit of this as n approaches infinity, that
also has to be equal to zero. So the limit as n approaches infinity of p over six to the nth power
also has to be equal to zero. Under what conditions is
that going to be true? Well, to meet either
one of those conditions, p over six has to be less than one. If p over six was equal to one, if, for example, p was six, well then we wouldn't be monotonically decreasing. Every term here would just be one. It would be one to the one,
one squared, on and on and on. And if p is greater than
six, well, then every time we multiply by p over six again, we would get a larger
number over and over again, and the limit for sure
would not be equal to zero. We could say p over six
needs to be less than one, and so multiply both sides by six and you get p needs to be less than six. And they told us, for what are
all the positive values of p? So we also know that p has
to be greater than zero. P is greater than zero and less than six, which is that choice right over here. Once again, we're not gonna
say less than or equal to six because if p was equal to six, this term is gonna be one to the
n, and so we're just gonna have, this would be one. It would be one minus one plus one, on and on and on and on forever. So definitely like that first choice.