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## Integral Calculus

### Course: Integral Calculus > Unit 5

Lesson 7: Alternating series test# Alternating series test

When a series alternates (plus, minus, plus, minus,...) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0.

## Want to join the conversation?

- From1:20to1:32- what is the second condition given for?

Isn't it clear, that if lim(b_n) = 0 when n -> infinity, then {b_n} should be decreasing sequence only? How this sequence could be not decreasing if the limit of b_n (when n -> infinity) is equal zero?(16 votes)- I can't really explain how any of this works, but I can give a counter-example: sin(pi*n/4)/n is not a decreasing series, but still tends to zero as n approaches infinity.(23 votes)

- The restrictions on this test seem redundant.

from0:53to1:33, Sal gives three restrictions on the series:

1) Bn ≥ 0 for all relevant n (namely positive integers n).

2) lim as Bn→∞ = 0

3) {Bn} is a decreasing sequence.

Don't the first two rules imply the third?(6 votes)- Consider the function f(n) = x*e^(-n). This will be our Bn.

What do you get at n= 0, n=1, n=2?

f(0) = 0,

f(1) = e^-1 = 0.37,

f(2) = 2/e^-2 = 0.27

So the first 3 terms of the sequence are: [0, 0.37, 0.27].

Notice the b2> b1 this function is not always decreasing! Yet all terms are greater than zero. And the limit as n→∞ = 0. Some functions like this rise a little bit, then fall back down as they go on. So the third rule is necessary.(13 votes)

- Does this test work for an increasing sequence?(3 votes)
- This test is used to determine if a series is converging. A series is the sum of the terms of a sequence (or perhaps more appropriately the limit of the partial sums).

This test is not applicable to a sequence.

Also, to use this test, the terms of the underlying sequence need to be alternating (moving from positive to negative to positive and so on). Therefore the sequence would not be increasing but rather 'oscillating' between positive and negative terms.(8 votes)

- At0:38-0:48: Why is it necessary that the alternating sign be expressed by either (-1)^n or (-1)^(n+1). Aren't there a whole slew of other ways to produced an alternating sign? What about (-1)^(n+2), (-1)^(n^2), or any polynomial exponent with odd integer coefficients?(3 votes)
- We are only talking about the form the series takes on. We know that it alternates, so the question is, is a negative term first, or a positive term. Given n goes from 1 to infinity, the first term of the (-1)^n series will be negative, and the first term of the (-1)^(n+1) series will be positive. That is all that is meant by the form of the series. Why make it any more complicated? It is an alternating series, either the first term is positive, or the first term is negative.(6 votes)

- What is an example of a series that fails the alternating series test but still converges, as Sal mentions at the end?(4 votes)
- At5:03He says that we can use other techniques like the limit comparison test, I tired using the limit comparison test but I can't figure out another functions that behaves like this function and that I can figure out It converges ... can someone please tell me how :)(3 votes)
- I don't think we can use the limit comparison test in this case, since the limit comparison require terms in both sequences greater that 0

Maybe he's saying the ratio test, the ratio test also involve limit, and can prove the series converges(2 votes)

- While attempting some practice problems, I couldn't get the correct answer, and this came up as a hint. "This series meets all the conditions for the alternating series test and hence it converges. However, since we can show that ∑n=1∞ n+1n2 diverges by using a comparison test with ∑n=1∞1n. Thus the series converges conditionally." I do not understand what this means, What are the "conditions"?(2 votes)
- I had a similar experience, I'm not sure the hints are properly done on some of these later topics..(4 votes)

- Is this test also called the Leibnitz test for alternating series??(3 votes)
- Yes it is the same! Some books/profs call it alternating and some call it the Leibnitz test.(1 vote)

- Can you use the ratio test on an alternating series?(3 votes)
- So I just tried the first practice problem in the "Alternating Series" challenge, and I found that one of the hints contradicts what Sal says in this video. At5:32, Sal says, "
*Now once again, if something*"**does not pass**the Alternating Series Test, that**does not necessarily mean that it diverges**; it just means that you couldn't use the Alternating Series Test to prove that it converges.

The problem hint, however, says, "...Since lim n→∞ an = 1 ≠ 0, the series**fails the Alternating Series Test**and therefore**diverges**."

Help? Can the Alternating Series Test be used to*prove*DIVERGENCE, or only conclusively prove CONVERGENCE?(3 votes)- The alternating series test only proves an alternating series converges and nothing about whether the series could/will diverge.

Was that the first hint given?

Do you have the URL of the question?(1 vote)

## Video transcript

- [Narrator] Let's now expose ourselves to another test of convergence, and that's the Alternating Series Test. I'll explain the Alternating Series Test and I'll apply it to an
actual series while I do it to make the... Explanation of the Alternating Series Test a little bit more concrete. Let's say that I have some
series, some infinite series. Let's say it goes from N equals
K to infinity of A sub N. Let's say I can write it as or I can rewrite A sub N. So let's say A sub N, I can write. So A sub N is equal to
negative one to the N, times B sub N or A sub N is equal to
negative one to the N plus one times B sub N where B sub N is greater
than or equal to zero for all the Ns we care about. So for all of these integer
Ns greater than or equal to K. If all of these things, if
all of these things are true and we know two more things, and we know number one, the
limit as N approaches infinity of B sub N is equal to zero. Number two, B sub N is
a decreasing sequence. Decreasing... Decreasing sequence. Then that lets us know that
the original infinite series, the original infinite series, is going to converge. So this might seem a little
bit abstract right now. Let's actually show, let's
use this with an actual series to make it a little bit more, a little bit more concrete. Let's say that I had the series, let's say I had the series
from N equals one to infinity of negative one to the N over N. We could write it out
just to make this series a little bit more concrete. When N is equal to one, this
is gonna be negative one to the one power. Actually, let's just
make this a little bit, let's make this a little
bit more interesting. Let's make this negative
one to the N plus one. When N is equal to one, this is gonna be negative
one squared over one which is gonna be one. Then when N is two, it's
gonna be negative one to the third power which is gonna be negative one half. So it's minus one half plus one third minus one fourth plus minus and it keeps going
on and on and on forever. Now, can we rewrite
this A sub N like this. Well sure. The negative one to the N plus one is actually explicitly called out. We can rewrite our A sub N, so let me do that. So negative, so A sub N which
is equal to negative one to the N plus one over N. This is clearly the same
thing as negative one to the N plus one times one over N which is, which we can then say this thing right over
here could be our B sub N. This right over here is our B sub N. We can verify that our
B sub N is going to be greater than or equal to zero
for all the Ns we care about. So our B sub N is equal to one over N. Clearly this is gonna be
greater than or equal to zero for any, for any positive N. Now what's the limit? As B sub N, What's the limit of B sub
N as N approaches infinity? The limit of, let me
just write one over N, one over N, as N approaches
infinity is going to be equal to zero. So we satisfied the first constraint. Then this is clearly a decreasing sequence as N increases the denominators
are going to increase. With a larger denominator, you're going to have a lower value. We can also say one
over N is a decreasing, decreasing sequence for the Ns that we care about. So this satifies, this
is satisfied as well. Based on that, this thing is always, this thing right over here is always greater than or equal to zero. The limit, as one over
N or as our B sub N, as N approaches infinity,
is going to be zero. It's a decreasing sequence. Therefore we can say
that our originial series actually converges. So N equals 1 to infinity of negative one to the N plus over N. And that's kind of interesting. Because we've already seen that if all of these were positive, if all of these terms were positive, we just have the Harmonic Series, and that one didn't converge. But this one did, putting these
negatives here do the trick. Actually we can prove this
one over here converges using other techniques. Maybe if we have time, actually in particular
the limit comparison test. I'll just throw that out
there in case you are curious. So this is a pretty powerful tool. It looks a little bit about
like that Divergence Test, but remember the
Divergence Test is really, is only useful if you want
to show something diverges. If the limit of, if
the limit of your terms do not approach zero, then you say okay, that
thing is going to diverge. This thing is useful because you can actually
prove convergence. Once again, if something does not pass the alternating series test, that does not necessarily
mean that it diverges. It just means that you couldn't use the Alternating Series Test
to prove that it converges.