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Integral Calculus (2017 edition)
Course: Integral Calculus (2017 edition) > Unit 11
Lesson 1: Basic convergence testsnth term divergence test
If the terms of an infinite series don't approach zero, the series must diverge. Learn more about this test in this video.
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Is it always "If lim(n->infinity)a_n ≠ 0. then the series diverges" Can't the series converge at a different value?(12 votes)
It's easy to mix up the terminology here. A sequence is a list of numbers, and a series is a sum of a list of numbers. The test here is saying that if the sequence (the list of numbers) doesn't approach zero as we go to infinity, then the series (the sum) diverges.
So, you're thinking of the sequence, and you're absolutely right that it can converge to a number other than zero. But then the series has to go to infinity, because we're adding the numbers in the sequence. For example, the sequence 2, 3/2, 5/4, 9/8, 17/16 . . . approaches 1 as we go to infinity (subtracting the next power of 1/2 each time). But that means the series (which is the sum of all these values) looks like 1 + 1 + 1 + 1 . . . as we go to infinity, and a sum of an infinite number of 1's is infinite. The only way the series (the sum) can converge is if the sequence (the numbers we're adding) approaches zero.(51 votes)
If the series fails at the divergence test (Σ a_n ≠ 0), what condition(s) will make it convergent or not?(14 votes)
I am not sure what you mean by "fail at the divergence test." The divergence test is a conditional if-then statement.
If the antecedent of the divergence test fails (i.e. the sequence does converge to zero) then the series may or may not converge. For example, Σ1/n is the famous harmonic series which diverges but Σ1/(n^2) converges by the p-series test (it converges to (pi^2)/6 for any curious minds).
If the antecedent of the divergence test is true (i.e. Σ a_n ≠ 0) then the series certainly diverges. There is no way to make a divergent series converge except by changing the terms of the sequence.
I hope this helps to clarify a little bit. Take care!(27 votes)
how come it will diverge if the result is already 1/3? Doesn't it mean that it convergence(9 votes)- You have fallen victim to the confusion between a Sequence and a Series.
Sequence=a string of numbers {1,3,5,7................}
Series=SUM of a sequence. 1+3+5+7+..........
The divergence test discussed in this video tests the series's divergence by seeing if the sequence converges. If the sequence has terms that go to infinity, then the series (because it is a sum) will have to add that infinity, causing it to diverge. The series that aren't shown to be divergent by this test do so because the sequence they are summing converges, leaving them freedom to converge or diverge.
So, when thinking about convergence/divergence, you almost always need to clarify whether you are talking about the sequence or the series, because they are different things.(28 votes)
So if the nth term of a serie as n goes to infinity is not zero, this series will definetly diverge ?
I 'm a bit scpetical here, because of this example : the infinite sum from n=0 to n=infinity : Σ (-1)^n
It looks like this : 1 - 1 + 1 - 1 + 1 - 1...
I think that the limit is not zero here, so it should diverge, however, I ve seen that this series does actually converge to 0.5
I m confused...
Can anyone explain me where it is wrong ? Thanks !(5 votes)
First of all, why are you more skeptical to the divergence test than to the summation of∑ (-1)ⁿ? I would argue that a result of1/2on the latter is less intuitive.
It is in fact true that the series∑ (-1)ⁿdiverges, since the numbers(-1)ⁿdo not tend to a limit asn → ∞(in particular, they do not converge to zero). Recall that convergence means that the sequence of partial sums tends to a limit. There are other ways to assign a number to infinite series, even divergent ones. One such method is called Cesáro summation, and it is in this sense that∑ (-1)ⁿ⁺¹sums to1/2. It is important to realise that this does not mean, at all, that the series converges to1/2(it is divergent) - it means that it has Cesáro sum equal to1/2, which is a number obtained by a completely different limiting process.
Cesáro summation is defined as follows: given a sequence{a(n)}of complex numbers, lets(N)denote theNth partial sum of the series∑ a(n), i.e., lets(N) = a(1) + a(2) + … + a(N).
Now consider the sequence{c(M)}defined byc(M) = (1/M)[s(1) + s(2) + … + s(M)],
i.e.,c(M)is the arithmetic mean of theMfirst partial sums of the series∑ a(n). If the sequence{c(M)}tends to a limit, call itC, asM → ∞, we say that the series∑ a(n)is Cesáro summable and has (Cesáro) sumC.
One can show that whenever the series∑ a(n)converges toA, then it is Cesáro summable and has Cesáro sumA. Moreover, there exist divergent series which are Cesáro summable. It is in this sense that Cesáro summability extends the notion of summation of infinite series. There are other summation methods as well.(8 votes)
Is the increasing value "n" of series always a natural number?
Are there series expressed using increasing values other than natural ones?(2 votes)
Sort of.
When n is used as the index number of a sequence or series, it has to be a nonnegative integer, yes. However, there is no requirement that members of a sequence or terms of a series have to be related to their index number by some mathematical process.
For example, the series of the prime numbers does not have a pattern where you could relate each term to its index number. However, we are mostly interested in sequences and series that can be determined by some mathematical operation.
If there is a need, say to increase by some value that is not an integer, you can always construct a function to describe how the subsequent terms vary that uses the n but then performs algebraic operations on it. For example, perhaps the terms are (⅔n+π)^(1/n) where n is the index number.
NOTE: When using the ∑ notation. one sometimes sees a non-integer or a negative integer used in the bounds, but this is rather usual. It is standard and usually possible to write the notation in such a way that the lower bound is a nonnegative integer. In any case, the index would increase by +1 as you go from the lower bound to the upper bound.(5 votes)
- Would (-1)^n be diverging? Because it's obviously not converging. The answer repeatedly oscillates between 1 and 0, and there is no limit as n approaches infinity. In the practice problems, (-1)^(n+1) was considered diverging because the limit as n approaches infinity does not equal 0. But it doesn't equal anything, just like sin. n or cos. n. How would this be defined?(3 votes)
I answered your other question regarding this concept; I'll just leave a note here, too. Because the nth term test deals with finite limits (or definitely infinite ones), it's not really helpful for a series likeSUM[ (-1)^n ]because we can't define the limit.(1 vote)
I fell like the nth term divergence test is a bit ambiguous because n is typically used for the final integer in the index and i is usually used for the index. I know a_n is a sequence is usually uses n, but nonetheless when I think of nth term I mistake this for the nth partial sum. Does anyone have a good justification for this?(1 vote)
The phrase "the nth term" just refers to the expression that allows the calculation of the term that is in the nth position.
You yourself say "n is typically used for the final integer". That means you admit that n is not always used as the last term. As you continue you will see a variety of letters used,
eg i=1 to k, n=1 to m, k=1 to p, etc.
My advice is to relax your notion that "n is typically used for the final integer in the index and i is usually used for the index" and be able to generalize the concept so you will not be hampered by which letters an author uses to describe a process.
I cannot overstate how important the ability to generalize notations/systems is to your future success in math.(5 votes)
- How would you find what Sum of 1/n^2 converges to?(1 vote)
With difficulty. This is known as the Basel problem - first solved by Euler in 1734. The answer is π²/6
See here: https://en.wikipedia.org/wiki/Basel_problem for proofs and approaches.(5 votes)
Because the divergent tests are not 100% percents efficient, when should we choose to use or not to use this kind of test?(1 vote)- The nth term divergence test ONLY shows divergence given a particular set of requirements. If this test is inconclusive, that is, if the limit of a_n IS equal to zero (a_n=0), then you need to use another test to determine the behavior. As you learn more tests, which ones to try first will become more intuitive.(4 votes)
- So even if this test comes back with the limit equaling zero, that does not mean the series converges?(1 vote)
The nth term test can only tell you that a series diverges, but cannot tell you that a series converges. If the nth term fails to approach 0, the series definitely diverges, but if the nth term does approach 0, the series could converge or could diverge.
An example of a series with nth term approaching 0 but that diverges is the harmonic series 1 + 1/2 + 1/3 + 1/4 + 1/5 + ...
Have a blessed, wonderful day!(2 votes)
Video transcript
- [Voiceover] What we're going to do now is start to explore a series of tests to determine whether a series
will converge or diverge and the first one I'm going
to go through right now is perhaps the most
basic and hopefully see the most intuitive and this
is the divergence test. The divergence test won't tell
us if a series will converge but it can tell us if something
will definitely diverge. First let me write it in
kind of mathy notation and then we'll look at an
actual concrete example on it. The divergence test tells
us that if the limit as N approaches infinity of
A sub N does not equal zero, then the infinite series
going from N equals one to infinity of A sub N will diverge. We've already gone through
what it means to diverge and this sum is either going to go unbounded to positive infinity or unbounded to negative infinity or it'll just oscillate between values, it'll never really approach
a given sum or given value. That's what the divergence test tells us and you're probably
thinking, okay all right, I can kind of get what this says but where is this actually useful? To see where it can be useful let's look at a candidate series and see if we can figure out
if it's going to diverge. Let's say I had this series so the sum from N equals one to infinity and in general it doesn't
always have to be N equals one, it could be N equals five, it could be N equals zero. The key is, is that this
is an infinite series that we're talking about
so that's why we care about the limit is N approaches infinity. Let's say we have the
sum of four N squared minus N to the third over seven, minus three and to the third power. Given what we know about
the divergence test, is this series going
to converge or diverge? Well, let's just look at this, what we're taking the sums of. So this is essentially
or this is our A sub N if we're trying to match to the definition or I guess the explanation
of the divergence test. Let's think about what the limit, as N approaches infinity of four N squared minus N to the third over seven minus three N to the third is. I encourage you to pause the
video and think about that. Well there's a couple of
ways to think about it, one way is, hey look as N goes to infinity the highest degree terms of the numerator and the denominator are the ones that matter. So this is going to approach
negative N to the third over negative three N to the third which will just be which
would approach negative one over negative three which
would be positive 1/3 or we could say if we want
to do a little bit more, or do a little bit more systematically. Limit as N approaches infinity, we can divide what the
numerator and the denominator by end of the third, so if we divide the
numerator by end of the third this first third is
going to be four over N minus one over seven over N
to the third, minus three. Here it becomes clear, the
limit is N approaches infinity, that's going to go to zero,
that's going to go to zero and so you're going to
be left with negative one over negative three which is equal to 1/3. Notice, the limit of A sub
N as N approaches infinity in this case is not equal to zero, therefore this sum will, this
infinite series will diverge. Now let's think for a second
why this makes a ton of sense. Well, the only way that
you're going to converge that you're going to set, remember you're taking infinite sum. You're taking an infinite sum of things so the only reasonable way that something might be able
to converge to a finite value is if every extra term you're
adding is getting smaller and smaller and smaller
as approaching zero. If as N approaches
infinity, you go unbounded or if it's even equal to 1/3, this means that for very large ends you just keep adding things
that are getting closer and closer to 1/3. Well if you add an infinite
number of one-thirds together, you're going to go to infinity, you're going to be unbounded. You are going to diverge so
that's all it's telling us. Look, in order for something to converge as you're taking infinite sum of them, at some point these things
are going to have to get really, really, really close to zero. If at some point they're
not getting close to zero, there's no way is going to converge, that thing will diverge so hopefully that makes a little sense. That also gives you I
guess another insight on what the divergence test can't do. The divergence test can be used to show that something will diverge but if something, I guess you could say passes the divergence test, it doesn't or I guess
fails the divergence test or if this isn't true, it doesn't mean that the
thing is going to converge. Let me give you an example of that. This right over here from
N equals one to infinity of one over N and this is
actually the harmonic series right over here. This, if we look at, if we try
to apply the divergence test, we would say, okay, well what's the limit as N approaches infinity of one over N. Well, hey that is zero, if
that approaches infinity this is equal to zero. We could say well it's kind
of failing the divergence test so we're not just by
using the divergence test we can't prove that this
thing is going to diverge but that doesn't mean
that it doesn't diverge. It actually turns out and we
prove this in several videos, it actually turns out that
this thing does diverge. This thing does diverge, it's just that the
divergence test isn't enough, it's not enough of a tool
to let us know for sure that this diverge, we'll
see the comparison test and the integral test can either be used to prove that this in fact does diverge. You can definitely not
say that if something, if this does not apply for something. If you think the limit
as N approaches infinity and it does go to zero,
that still might diverge. It doesn't necessarily
mean that it converges. Now there are things that do converge that where they do approach zero so for example if I take the sum from N equals one to infinity
of one over N squared. If I try to apply the divergence test, I have the limit as N approaches infinity of one over N squared,
well this does equal zero. This gets really, really large, this thing is going to approach zero and so once again this is I guess failing the test for divergence. Just with that alone
you don't know for sure that this thing is going to converge, it still might diverge, the divergence test just
might not have been enough. Now it does turn out and
once again we prove this in future videos that
this thing does converge. This thing does converge
but not because it, I guess you could say
fails the divergence test. We have to use a different test to show that this does in fact converge, just as we have to use
a different test to show that this does in fact diverge. Where the divergence test is useful is for the things that actually
pass the divergence test. When you actually find that
the limit is N approaches infinity of A sub N does not equal zero, like this case right over here. In this case the divergence test helps us because it helps us make the conclusion that this series definitely diverges.