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Infinite series as limit of partial sums

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.1 (EK)
,
LIM‑7.A.2 (EK)
Infinite series are defined as the limit of the infinite sequence of partial sums. Since we already know how to work with limits of sequences, this definition is really useful.

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  • starky ultimate style avatar for user Marko Arezina
    How do you come up with the formula for partial sums of non geometric series? The limit strategy is good to evaluate the sum but it all seems kind of useless if you cannot come up with the formula.
    (25 votes)
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    • aqualine ultimate style avatar for user awesomediabrine
      There are several ways. One is to use algebra to deduce a formula, like Gauss did with S = 1,2,3,4,...(http://mathandmultimedia.com/2010/09/15/sum-first-n-positive-integers/). Another is to use established partial sums to derive new ones, like a generalization of Gauss' method to any arithmetic series. A third way is to use induction. For example, given the series 1,3,5, ..., the partial sum looks a lot like n^2 as one does S1, S2, etc. You can then prove this inductively. First the series is 2n+1. We want to prove that n^2 = S_n, so plugging in n we see that n^2=n^2, therefore the next partial sum is the next term(2n+1) + the sum of the pervious n terms (n^2). Plugging in the next n into our partial sum formula we see that (n+1)^2 = n^+2n+1, which is what we got earlier. This shows that given a partial sum = n^2, all partial sums after that follows that pattern. Then we simply do 1+3 = 2^2 to prove that there is a partial sum = n^2. I imagine that there are more ways but those are the ways I can think of. I feel like on AP tests they will be given though.
      (1 vote)
  • blobby green style avatar for user Sue Blaisdell
    In this video the teacher refers to a term and its subscript (S sub 3 for example) as the sum of the first 3 terms, but in the next video each term and its subscript are evaluated separately for a value. Which is true or is one true sometimes and not other times?
    (7 votes)
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    • male robot hal style avatar for user Yamanqui García Rosales
      Depends on what S means. If you are talking about sequences and S represents a sequence, then Sᵢ will refer to the value of the i-th element in the sequence.

      If you are talking about series, and S represents the partial sum of the series, then Sᵢ represents the partial sum up to the i-th item of the series.

      There is nothing special about the letter S, you can use any symbol you want to represent either sequences or series.
      (9 votes)
  • aqualine tree style avatar for user Evelyn Jung
    What are the potential errors that arise if, when testing for convergence, we take the limit of the general formula of the last element in the sequence instead of the partial sum?
    (4 votes)
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    • mr pink red style avatar for user andrewp18
      Good question! Let's think about this.
      Let 𝑎(𝑛) denote the 𝑛th term in a series and let 𝑆(𝑛) denote the 𝑛th partial sum. we know that if lim(𝑛 → ∞) 𝑆(𝑛) converges, then the infinite sum exists and is equal to that limit. However, what if lim(𝑛 → ∞) 𝑎(𝑛) converges? What does this say about 𝑆(𝑛)? Well we know that:
      𝑆(𝑛 + 1) = 𝑎(𝑛) + 𝑆(𝑛)
      We can take the limit of both sides:
      lim(𝑛 → ∞) 𝑆(𝑛 + 1) = lim(𝑛 → ∞) [𝑎(𝑛) + 𝑆(𝑛)]
      Suppose that the infinite series exists at 𝑘 (i.e. lim(𝑛 → ∞) 𝑆(𝑛) = 𝑘). Then we have:
      𝑘 = 𝑘 + lim(𝑛 → ∞) 𝑎(𝑛) ⟹ lim(𝑛 → ∞) 𝑎(𝑛) = 0
      So the sum can only converge if the terms are approaching 0. This makes sense intuitively because the terms need to get smaller and smaller if we have any hope of our sum "jumping" around less and less and "settling" on a single value. If the terms converged to any nonzero number, then naturally, the sum diverges as the sum would keep going on and on.

      But it should be noted that this is a necessary but not a sufficient condition for series convergence. It is possible for the terms of a series to converge to 0 but have the series diverge anyway. The classic example of this is the harmonic series:
      𝚺(𝑛 = 1) ^ ∞ [1/𝑛]
      Obviously here, the terms approach 0, (lim(𝑛 → ∞) 1/𝑛 = 0) but in fact, this sum diverges! So the fact that the terms of a series approach 0 is a necessary but insufficient condition for series convergence. On the other hand, the fact that the partial sums of a series converge is in fact a sufficient condition for convergence because this is exactly what we define series convergence to be. An infinite sum exists iff the sequence of its partial sums converges.

      Comment if you have questions!
      (2 votes)
  • starky ultimate style avatar for user snowrobin012
    At why does Sal divide, both the numerator and denominator by n^2?
    (2 votes)
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    • female robot grace style avatar for user loumast17
      It is a method to more easily see what the end behavior will be. Hopefully you see that dividing the numerator and denominator by the same thing doesn't change the fraction.

      Now, when you divide by the highest degree variable you get a bunch of terms with variables only in the denominators. When you look at a term that has a variable only in the denominator and its end behavior that term goes to 0, since one over a very large number is practically 0, and the larger ti gets the closer to 0 it gets.

      This means only the terms that had degrees to match the largest degree you divided by are left. this means the function, whatever it is, becomes identical to a function with just those terms.

      An example, if you look at (x^2 + x + 1)/(x^2 + 10000000x + 1000000000), if you compared the two at higher and higher x values they would look more and more like the same function. Specifically x^2 / x^2 = 1
      (2 votes)
  • blobby green style avatar for user Jillian Zhu
    Shouldn't he have divided the n values by n^3 instead of n^2 because n^3 is the highest exponent?
    (2 votes)
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  • ohnoes default style avatar for user Bhavana Sanjay
    heyy, what does s base infinity equals to?
    (2 votes)
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  • male robot johnny style avatar for user Alex Lee
    Aren't there are nested sums with infinite limits?
    (2 votes)
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  • leaf blue style avatar for user Katieashleysullivan
    If the limit of the partial sums is 0, does that mean that the sum of the series equals 0?
    (2 votes)
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    • mr pants teal style avatar for user sauj123
      Yes. because the sum of the series is defined as the limit of partial sum sub n with n approaching towards infinity. As far as I know (if I'm wrong please correct me), this can only happen if either a sub n oscillates around zero with positive and negative numbers of the same magnitude somewhere in the sequence (e.g. 5 - 7 - 3 + 7 - 5 + 3 + ... = 3 - 3 + 5 - 5 + 7 - 7 + ... = 0), or the partial sum always stays at zero as a sub n is also equal to zero.
      (1 vote)
  • piceratops ultimate style avatar for user Michael William Clayton
    In the lesson "Worked example: sequence convergence / divergence," the fourth example was that wonderful sequence d_n = (-1)^n, the terms of which oscillate between 1 and -1. The terms of the partial series are interesting: {1, 0, 1, 0, 1 ...} ... so these oscillate as well.

    My question, then, is: how can we use limit analysis to confirm that, according to convention, this oscillating set of partial sums also diverges?
    (1 vote)
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    • leaf green style avatar for user kubleeka
      That's just a matter of showing that the sequence {1, 0, 1, 0...} diverges. If the limit of the sequence was L, then there would be a point in the sequence after which every term was within, say, 1/2 of L.

      Every term of the sequence is either 0 or 1, so we need an L such that |1-L|<1/2 and |0-L|=|L|<1/2.
      But this first inequality means that L is in the interval (1/2, 3/2) and the second one means that L is in (-1/2, 1/2). These intervals don't intersect, so no such L exists.

      Since the limit doesn't exist, the sequence is divergent.
      (2 votes)
  • leaf green style avatar for user Amandeep Singh
    If the limit to infinity of the partial sum is a finite value, then is that finite value also the sum of the series?
    (1 vote)
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    • blobby green style avatar for user Chuck Finley
      Yes. If the limit of the partial sums exists - is a finite value - then the series converges and the series equals the limit. Also see the answer below by sauj123, who answered with respect to the specific case of the limit being zero. Consider his reminder of the definition of an infinite series.
      (1 vote)

Video transcript

- [Voiceover] Let's say that we have an infinite series S so that's the sum from n = 1 to infinity of a sub n. We could write it out a sub 1 plus a sub 2 and we're just going to go on and on and on for infinity. We're going to go on and on and on forever. So, let's say, and I've written it in very general terms let's say we have a formula for the partial sums of S. We know that S sub n is equal to 2n to the third over n plus 1 times n plus 2. Now, my question to you is, based on what I've just told you S is the sum in a very general way written this infinite series but I have the partial sum. The sum of the first n terms of S is given by this formula right over here does this series converge or diverge? Does this thing converge to some finite value or is it unbounded and does it diverge? Well, one way to think about this is the idea that our infinite series S is just the limit as n approaches infinity of our partial sums. So, what do we mean by that? Well, you could a sequence of partial sums here. You have S sub 1, S sub 2, S sub 3 and you keep going so this would be the sum of the first term. This would be the sum of the first 2 terms. This would be the sum of the first 3 terms and just think about what happens to this sequence as n right over here approaches infinity because that's what this series is. It's the sum of the first, I guess you could say the first, infinite terms. It's the sum of all, you have an infinite number of terms here. Well, let's think about what this. The limit is n approaches infinity of S sub n. That's just going to be the limit as n approaches infinity of this business right over here. 2n to the third power over n plus 1 times n plus 2 and there's several ways you could evaluate this. One way is you could just realize, "Hey, look in the bottom this is going "to be a second degree polynomial." On up here, you have a third degree so the numerator is gonna grow faster than the denominator so this is going to be unbounded. So that will immediately tell you well this is gonna approach infinity so S is going to diverge but if you wanna do it a little bit less hand wavy than that we can actually do a little bit more algebra. Limit as n approaches infinity, 2n to the third power over, let's multiply this out, n squared plus 3n plus 2 and, let's see, we can divide the numerator and the denominator by n squared. So, this is going to be the limit as n approaches infinity of, if we divide the numerator by n squared, you're going to have, actually, let's divide the numerator and, well yeah let's divide it by n squared so, if we divide the numerator by n squared, we're gonna have 2n and then the denominator divided by n squared you're gonna have 1 plus 3 over n plus 2 over n squared. Now, when you look at it like this, it becomes pretty clear this thing as n approaches infinity, this thing is gonna towards infinity but this thing down here the denominator this is gonna go towards 0. This is gonna go towards 0 so the denominator's gonna go towards 1. So, this whole thing, is the limit is gonna go to infinity and since the limit of the partial sums goes to infinity that mean that this infinite series is not going to be a finite value. It's just going to diverge. So, this character right over here is going to diverge. In order for it to have converged, this thing should have come, this limit should have been some finite value. So, hopefully, that makes sense. All we say is, "Look, infinite series, "we had a formula for the partial sum "of the first n terms and then we said oh look "the series itself, the infinite series, "you could view it as a limit of, "as n approaches infinity, of the partial sum "S sub n and we said hey, that approach infinity "this thing is diverging."