If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

nth-term test

# nth term divergence test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.5 (EK)

## Want to join the conversation?

• If the series fails at the divergence test (`Σ a_n ≠ 0`), what condition(s) will make it convergent or not?
• 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!
• Is it always "If lim(n->infinity)a_n ≠ 0. then the series diverges" Can't the series converge at a different value?
• 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.
• how come it will diverge if the result is already 1/3? Doesn't it mean that it convergence
• 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.
• 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 !
• 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 of `1/2` on the latter is less intuitive.

It is in fact true that the series `∑ (-1)ⁿ` diverges, since the numbers `(-1)ⁿ` do not tend to a limit as `n → ∞` (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⁺¹` sums to `1/2`. It is important to realise that this does not mean, at all, that the series converges to `1/2` (it is divergent) - it means that it has Cesáro sum equal to `1/2`, 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, let `s(N)` denote the `N`th partial sum of the series `∑ a(n)`, i.e., let

`s(N) = a(1) + a(2) + … + a(N).`

Now consider the sequence `{c(M)}` defined by

`c(M) = (1/M)[s(1) + s(2) + … + s(M)],`

i.e., `c(M)` is the arithmetic mean of the `M` first partial sums of the series `∑ a(n)`. If the sequence `{c(M)}` tends to a limit, call it `C`, as `M → ∞`, we say that the series `∑ a(n)` is Cesáro summable and has (Cesáro) sum `C`.

One can show that whenever the series `∑ a(n)` converges to `A`, then it is Cesáro summable and has Cesáro sum `A`. 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.
• Is the increasing value "n" of series always a natural number?
Are there series expressed using increasing values other than natural ones?
• 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.
• 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?
• 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 like `SUM[ (-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.
• How would you find what Sum of 1/n^2 converges to?
(1 vote)
• 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.