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

Convergent and divergent sequences

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.1 (EK)
,
LIM‑7.A.2 (EK)
A sequence is "converging" if its terms approach a specific value as we progress through them to infinity. Get an intuitive sense of what that even means! Created by Sal Khan.

Want to join the conversation?

  • spunky sam blue style avatar for user Chunmun
    What does diverges or converges means here ?
    Please help.
    Thank you.
    (19 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Christi
      Converging means something is approaching something. Diverging means it is going away. So if a group of people are converging on a party they are coming (not necessarily from the same place) and all going to the party. Similarly, for functions if the function value is going toward a number as the x values get closer, then the function values are converging on that value.
      (58 votes)
  • blobby green style avatar for user darek
    At , why is it (-1)^(n+1)? Is that the same as (1)^(n-1)?
    Thanks
    (14 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user AJ
      (-1)^(n-1) does equal (-1)^(n+1) for every n value. Both equations oscillate every other N, every (-1)^ to an even number equal(-1)^ to a different even number. You can change the (n-1) to any odd number [for example (n+67) or (n-7)] and get the same answer.
      (18 votes)
  • piceratops ultimate style avatar for user Ian Fisher
    I thought 1 to the infinite power was an indeterminate form, so don't you need to use L'Hopital's rule to evaluate the limit in the video?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Creeksider
      Suppose we want to know the limit of a^b as x goes to infinity, where a and b are both functions of x. If we find that a approaches 1 and b approaches infinity, we have an indeterminate form, because we can't tell without further analysis whether the forces attracting a toward 1 (making the expression approach 1) are overpowered by the forces moving b toward infinity (making the expression approach infinity or zero, depending on whether a is slightly greater than or less than 1). So that is why we say 1^∞ is an indeterminate form. HOWEVER, if a is not some function that approaches 1, but is actually the number 1, then we no longer have an indeterminate form. The expression 1^b is always 1, no matter how large or small the exponent. We no longer have an infinitesimal increment away from 1 that can be overpowered by the increase of the exponent. So we have an indeterminate form when we have a base approaching 1 and exponent approaching infinity, but not when we have a base that EQUALS 1 and exponent approaching infinity.
      (18 votes)
  • blobby green style avatar for user Doris Ayala
    what is the difference between this statements:
    a) The sequence of Bn has no limit.
    b) The sequence of Bn diverges to positive infinity.
    c) The sequence of Bn is simply divergent.
    d) limBn=(symbol of infinity)
    n--->(symbol of infinity)
    (5 votes)
    Default Khan Academy avatar avatar for user
    • leaf grey style avatar for user Qeeko
      I'll assume {B(n)} is a sequence of real numbers (but a sequence in an arbitrary metric space would be just as fine).

      a) {B(n)} has no limit means that there is no number b such that lim (n→∞) B(n) = b (this may be cast in terms of an epsilon type of definition).

      b) That {B(n)} diverges to +∞ means that for every real number M there exists a real number N such that B(n) ≥ M whenever n ≥ N.

      c) A sequence is divergent if and only if it is not convergent, hence this means the same as a).

      d) This means the same as b).
      (5 votes)
  • blobby green style avatar for user Lindani Maqetuka
    What is the difference between convergent sequence and a converging series?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf grey style avatar for user Jason Hoff
    is this a harmonic series?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • orange juice squid orange style avatar for user Selina Cox
    don't all functions approach a number? like isn't infinity a number? cone somebody give me an example of a function that diverges.
    (0 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user davis
      No. Not all functions approach a number as their input approaches infinity. One of the main things a function has to do to approach a number is to start to stabilize. Take sine or cosine. We know they will never output anything greater than 1, or less than -1, we are even able to compute them for any real number. But, we know that they will always fluctuate. They don't head to infinity, and they don't converge. If we were to investigate sin(x)/x, it would converge at 0, because the dividing by x heads to 0, and the +/- 1 can't stop it's approach.
      A similar resistance to staying mostly still can be found in equations that diverge as their inputs approach infinity. What number does 2^x go to? (It diverges)
      We might claim that it goes to 34359738368, a very big number equal to 2^35, but we 2^x, being fickle to the point of cruelty, will leave this value (already large) for a notably larger one as x goes to 36. And remember, infinity is MUCH larger than 36, and this function will continue to leave potential values in the dust, never looking back, indefinitely.
      Also, just in case it isn't clear, infinity doesn't follow many of the properties that numbers follow. infinity + 1 = infinity, infinity^2 = infinity / 8, basically infinity can't be dealt with in algebra, (although L'Hopitol's rule lets us manage it), and shouldn't be thought of as a "number" in the common sense of the word.
      (2 votes)
  • aqualine sapling style avatar for user ioakim.koutsioukis
    why do we use n+1 instead of n-1 when defining the power of (-1)? is it just semantics or common protocol? I know in other cases it would be n+1 is there any reason for it not to be n-1?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • starky sapling style avatar for user Albaraa Kamal
    What is the difference between finding the limit of a sequence and finding the limit of a function?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Muhammad Ali Siddiqui
    sequence (1+(-1)^n) as n goes to infinity is divergent, but how to prove that
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

Let's say I've got a sequence. It starts at 1, then let's say it goes to negative 1/2. Then it goes to positive 1/3. Then it goes to negative 1/4. Then it goes to positive 1/5. And it just keeps going on and on and on like this. And we could graph it. Let me draw our vertical axis. So I'll graph this as our y-axis. And I'm going to graph y is equal to a sub n. And let's make this our horizontal axis where I'm going to plot our n's. So this right over here is our n's. And this is, let's say this right over here is positive 1. This right over here is negative 1. This would be negative 1/2. This would be positive 1/2. And I'm not drawing the vertical and horizontal axes at the same scale, just so that we can kind of visualize this properly. But let's say this is 1, 2, 3, 4, 5, and I could keep going on and on and on. So we see here that when n is equal to 1, a sub n is equal to 1. So this is right over there. So when n is equal to 1, a sub n is equal to 1. So this is y is equal to a sub n. Now, when n is equal to 2, we have a sub n is equal to negative 1/2. When n is equal to 3, a sub n is equal to 1/3, which is right about there. When n is equal to 4, a sub n is equal to negative 1/4, which is right about there. And then when n is equal to 5, a sub n is equal to positive 1/5, which is maybe right over there. And we keep going on and on and on. So you see the points, they kind of jump around, but they seem to be getting closer and closer and closer to 0. Which would make us ask a very natural question-- what happens to a sub n as n approaches infinity? Or another way of saying that is, what is the limit-- let me do this in a new color-- of a sub n as n approaches infinity? Well, let's think about if we can define a sub n explicitly. So we can define this sequence as a sub n where n starts at 1 and goes to infinity with a sub n equaling-- what does it equal? Well, if we ignore sign for a second, it looks like it's just 1 over n. But then we seem like we oscillate in signs. We start with a positive, then a negative, positive, negative. So we could multiply this times negative 1 to the-- let's see. If we multiply it times negative 1 to the n, then this one would be negative and this would be positive. But we don't want it that way. We want the first term to be positive. So we say negative 1 to the n plus 1 power. And you can verify this works. When n is equal to 1, you have 1 times negative 1 squared, which is just 1, and it'll work for all the rest. So we could write this as equaling negative 1 to the n plus 1 power over n. And so asking what the limit of a sub n as n approaches infinity is equivalent to asking what is the limit of negative 1 to the n plus 1 power over n as n approaches infinity is going to be equal to? Remember, a sub n, this is just a function of n. It's a function where we're limited right over here to positive integers as our domain. But this is still just a limit as something approaches infinity. And I haven't rigorously defined it yet, but you can conceptualize what's going on here. As n approaches infinity, the numerator is going to oscillate between positive and negative 1, but this denominator is just going to get bigger and bigger and bigger and bigger. So we're going to get really, really, really, really small numbers. And so this thing right over here is going to approach 0. Now, I have not proved this to you yet. I'm just claiming that this is true. But if this is true-- so let me write this down. If true, if the limit of a sub n as n approaches infinity is 0, then we can say that a sub n converges to 0. That's another way of saying this right over here. If it didn't, if the limit as n approaches infinity didn't go to some value right here-- and I haven't rigorously defined how we define a limit-- but if this was not true, if we could not set some limit-- it doesn't have to be equal to 0. As long as it-- if this was not equal to some number, then we would say that a sub n diverges.