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.

Main content

Worked example: sequence convergence/divergence

LIM‑7 (EU)
LIM‑7.A (LO)
LIM‑7.A.1 (EK)
LIM‑7.A.2 (EK)
How can we tell if a sequence converges or diverges? See Sal in action, determining the convergence/divergence of several sequences. Created by Sal Khan.

Want to join the conversation?

  • piceratops ultimate style avatar for user Robert Checco
    I am confused how at Sal is sure that is converges. I understand that the N^2 is going to grow at the faster rate, but wont the numerator become larger and larger than the denominator? In the denominator as "n" gets larger and larger the 10n is being subtracted from n^2. Whereas on the top the "n" is recieving addition by 9n. At what point is the "10n" or the "8n" enough to change the limiting value.
    (15 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Creeksider
      The key is that the absolute size of 10n doesn't matter; what matters is its size relative to n^2. When n=100, n^2 is 10,000 and 10n is 1,000, which is 1/10 as large. When n=1,000, n^2 is 1,000,000 and 10n is 10,000. We increased 10n by a factor of 10, but its significance in computing the value of the fraction dwindled because it's now only 1/100 as large as n^2. Each time we add a zero to n, we multiply 10n by another 10 but multiply n^2 by another 100. Eventually 10n becomes a microscopic fraction of n^2, contributing almost nothing to the value of the fraction.
      (67 votes)
  • marcimus pink style avatar for user Oya Afify
    if i had a non convergent seq. that's mean it's divergent ?
    (11 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Oskars Sjomkāns
    So if a series doesnt diverge it converges and vice versa? Is there no in between? And why does the C example diverge? It converges to n i think because if the number is huge you basically get n^2/n which is closer and closer to n.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      There is no in-between. All series either converge or do not converge. By definition, a series that does not converge is said to diverge.

      However, not all divergent series tend toward positive or negative infinity. Some series oscillate without ever approaching a single value.

      Now, there is a special kind of convergent series called a "conditionally convergent series". In this type of series half of its terms diverge to positive infinity and half of them diverge to negative infinity; however, the overall sum actually converges to some number.

      An example of a conditionally convergent series is:
      ∑ n=1 to infinity of { (-1)^(n+1)/(ln(8)*n)}
      This converges to ⅓. However, its negative terms diverge to negative infinity and its positive terms diverge to positive infinity.
      (20 votes)
  • aqualine sapling style avatar for user idkwhat
    Why does the first equation converge? I thought that the limit had to approach 0, not 1 to converge?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • primosaur ultimate style avatar for user Derek M.
      I think you are confusing sequences with series. Remember that a sequence is like a list of numbers, while a series is a sum of that list. Notice that a sequence converges if the limit as n approaches infinity of An equals a constant number, like 0, 1, pi, or -33. However, if that limit goes to +-infinity, then the sequence is divergent. If the first equation were put into a summation, from 11 to infinity (note that n is starting at 11 to avoid a 0 in the denominator), then yes it would diverge, by the test for divergence, as that limit goes to 1. However, since it is only a sequence, it converges, because the terms in the sequence converge on the number 1, rather than a sum, in which you would eventually just be saying 1+1+1+1+1+1+1...
      (6 votes)
  • male robot johnny style avatar for user Jayesh Swami
    In the option D) Sal says that it is a divergent sequence......
    Lets assume S = 1-1+1-1+1-1.... then
    1 - S = 1- (1-1+1-1+1-1+1...) which is 1-S = 1-1+1-1+1-1... which is same as S
    1-S =S ..... S= 1/2 ..... doesn't that means that it is convergent if we get a specific value......
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      You cannot assume the associative property applies to an infinite series, because it may or may not hold. And, in this case it does not hold.
      In short, S fails to exist for a divergent series, thus computations with S are meaningless. They have no more meaning than the "proofs" 1=2 which contain a hidden division by zero.

      Let me add a counter-example.
      S = 1-1+1-1+1-1....
      Notice that if I add another copy of the sum 1-1+1-1+1-1.... at the beginning of the
      sum, I get exactly the same sum. It is still 1-1+1-1+1-1....
      S + S = S
      2S = S
      And since you "proved" S = ½
      It must be the case that
      2S = ½
      Thus, S = ¼
      And, of course, I can add as many sets of S to each other as a like and they will still be the same sum, they will still be 1-1+1-1+1-1.... So let us say that I added S to itself 999 times, giving me 1000S = S.
      Since S = ½ and S = ¼
      And since S = 1000S
      1000S = ½. Thus, S = 1/2000
      And 1000S = ¼. Thus S =1/4000
      This is obviously absurd and self-contradictory. Thus, we know that the math is wrong. It Is NOT the case that S=½ nor any of the other values we could come up with. It is not the case that the associative property holds for this particular series. And, for that matter, it does not hold that S + S = 2S.
      Since it is possible to have multiple contradictory sums for S, it must be the case that S fails to exist.
      Thus when S fails to exist, it is possible to get various nonsensical and contradictory solutions.

      Basic mathematical operations all require that S exists, if S does not exist the operations can still produce "answers" but they will be nonsense.

      BTW, the Numberphile video where they "proved" that S of the positive integers was -1/12 made use of such nonsense with divergent series. Their "proof" was utter nonsense.
      (5 votes)
  • blobby green style avatar for user Ahmed Rateb
    what is exactly meant by a conditionally convergent sequence ?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      It is a series, not a sequence.
      A series is defined to be conditionally convergent if and only if it meets ALL of these requirements:
      1. It is an infinite series.
      2. The series is convergent, that is it approaches a finite sum.
      3. It has both positive and negative terms.
      4. The sum of its positive terms diverges to positive infinity.
      5. The sum of its negative terms diverges to negative infinity.

      The commutative and associative properties do not hold for conditionally convergent series. Thus, it is possible (by using the associative property and/or the commutative property) to group the terms of a conditionally convergent series to make it look like the series converges to any arbitrarily chosen number or to make it look like the the series diverges. Thus, conditionally convergent series are quite difficult to work with and it is very easy to get nonsense answers that look like they are correct.

      In other words, a conditionally convergent series has the property that you get different answers if you rearrange or regroup the terms.
      (4 votes)
  • piceratops ultimate style avatar for user David Procházka
    At Sal says that the exponential grows much faster than the polynomial, and I kinda see that, but is there a video somewhere that proves this, or for example if i had 10^100*en in the denominator it would still diverge right?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Creeksider
      Assuming you meant to write "it would still diverge," then the answer is yes. As x goes to infinity, the exponential function grows faster than any polynomial. Not sure where Sal covers this, but one fairly simple proof uses l'Hospital's rule to evaluate a fraction e^x/polynomial, (it can be any polynomial whatever in the denominator) which is infinity/infinity as x goes to infinity. Repeated application of l'Hospital's rule will eventually reduce the polynomial to a constant, while the numerator remains e^x, so you end up with infinity/constant which shows the expression diverges no matter what the polynomial is.
      (3 votes)
  • piceratops ultimate style avatar for user Daniel Santos
    Is there any videos of this topic but with factorials? I need to understand that. I found a few in the pre-calculus area but I don't think it was that deep. Any suggestions?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Akshaj Jumde
    The crux of this video is that if lim(x tends to infinity) exists then the series is convergent and if it does not exist the series is divergent. Am I right or wrong ?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user elloviee10
    I thought that the first one diverges because it doesn't satisfy the nth term test?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user doctorfoxphd
      Don't forget that this is a sequence, and it converges if, as the number of terms becomes very large, the values in the sequence approach some number. It is kind of interesting to look at some of the close-in values. At n = 3, the value is -3.12. At n = 9 it is -18.89 At n = almost 10, it is -∞ Just after n=10, it is coming back from + ∞
      Then it starts to behave:
      n     a_n
      15 4.9067
      20 2.94
      100 1.212
      200 1.1002
      2000 1.0095

      So the values are getting closer and closer to 1, and as a sequence it converges
      But as a series, it definitely diverges
      All those ones summed together from an infinity of terms will get to infinity, not to mention the mess around n=10
      (3 votes)

Video transcript

So we've explicitly defined four different sequences here. And what I want you to think about is whether these sequences converge or diverge. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. And diverge means that it's not approaching some value. So let's look at this. And I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, the denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. And we care about the degree because we want to see, look, is the numerator growing faster than the denominator? In which case this thing is going to go to infinity and this thing's going to diverge. Or is maybe the denominator growing faster, in which case this might converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator and figure that out. So n times n is n squared. n times 1 is 1n, plus 8n is 9n. And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. The denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large, what dominates in the numerator-- this term is going to represent most of the value. And this term is going to represent most of the value, as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say I'm not rigorously proving it over here. But the giveaway is that we have the same degree in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100-- that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here. Well, we have a higher degree term. We have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster than the denominator. Or another way to think about it, the limit as n approaches infinity is going to be infinity. This thing's going to go to infinity. Now let's think about this right over here. So as we increase n-- so we could even think about what the sequence looks like. When n is 0, negative 1 to the 0 is 1. When n is 1, it's going to be negative 1. When n is 2, it's going to be 1. And so this thing is just going to keep oscillating between negative 1 and 1. So it's not unbounded. It's not going to go to infinity or negative infinity or something like that. But it just oscillates between these two values. So it doesn't converge to one particular value. So even though this one isn't unbounded-- it doesn't go to infinity-- this one still diverges. It doesn't go to one value. So let me write that down. This one diverges.