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Partial sums intro

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.1 (EK)
,
LIM‑7.A.2 (EK)
The partial sum of an infinite series as the sum of the first few terms (and hence it's only partial). This seemingly simple concept is very useful in thinking about infinite series.

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  • male robot johnny style avatar for user dennis
    At the end, isn't the partial sum a1+a2+a3+a4+a5+a6? Why did he only find a6 and call it good?
    (22 votes)
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    • orange juice squid orange style avatar for user rsr
      The partial sum is certainly that. However, we are given the formula for the sum. So, instead of writing a1 + a2 + a3 + a4 + a5 + a6, we could write Sn. Since we're looking for the partial sum of the first six, we can just plug in 6 into Sn and find the actual sum of the first six terms.
      (16 votes)
  • spunky sam red style avatar for user Satyam Sharma
    At , isn't that the Riemann's zeta function?
    (13 votes)
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  • starky ultimate style avatar for user Chris Bruno
    Just to be sure, we should have actually added a1 through a6 in the final example, right? in the S sub 3 example, we added 1 + 1/4 + 1/9, so we should have done the same for the S sub 6 problem as well, or am I just being crazy.
    (2 votes)
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    • leafers seed style avatar for user Travis Bartholome
      The two examples are a bit different. In the first, we know the general term for the series; that is, we can find a_n, but we aren't given a formula for S_n. So to find the sum, we add up all of the individual terms from a_1 to a_3 to get S_3.

      In the second example, we're given a formula for S_n, not a_n; so to get S_6, all we need to do is substitute 6 for n in the given formula, and we get the partial sum -- because we're given a formula for the sum this time. The formula tells us what the answer will be if we add up all the terms from a_1 to a_6; however, we don't have an easy way to determine what each individual term will be in this example, which is why Sal doesn't go through the process of adding terms manually like he did before.

      In short, in the first problem we have a formula to find each term but not the partial sum, whereas in the second problem we have a formula to find the partial sum but not each term.

      I hope that helps.
      (14 votes)
  • purple pi purple style avatar for user Jesse Wang
    How are partial sums different from finite series?
    (5 votes)
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  • female robot grace style avatar for user Anna
    How can you find the infinite sum of a converging series without doing infinitely many operations(which would take infinite time which is longer than our universe will last)
    (2 votes)
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    • piceratops ultimate style avatar for user Just Keith
      You can find the infinite sum if there is a pattern that is clearly followed which will inevitably lead to a particular sum as the number of terms approaches infinity.
      For example,
      ∑ 3/10ⁿ over n=1 to ∞
      The first few partial sums are:
      0.3
      0.33
      0.333
      0.3333
      And it is clear this pattern will continue forever. Thus, without having to keeping doing the sums, we know that as n approaches infinity, thus series approaches ⅓.

      Of course, it is not always that easy. There are some infinite series mathematicians have had to work a long time on. But the principle remains: if there is a clear pattern to the partial sums and it is clear that the pattern will continue forever, then if the sums are clearly converging to a single number, then that is the infinite sum.
      (8 votes)
  • duskpin ultimate style avatar for user Sudhanshu Basu Roy
    Sal, at , can't I simplify 33/220 and right it as 3/20? Why would be wrong in doing so?
    (4 votes)
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  • aqualine ultimate style avatar for user Insatiable
    Even though I see how partial sums work, I still want to know how to even obtain the formula for a particular partial sum. For example, if we have an infinite geometric series that starts as:
    1+2+4+8+...
    what would be the formula for the partial sum of the first n terms?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      (a(1-r^n))/(1-r) where a is the starting term, r is the ratio, basically what is being multiplied and n is the number term we are adding to.

      So in your example a = 1, r = 2 and n = 4 (1(1-2^4))/(1-2) = -15/-1 = 16 and you can check 1+2+4+8 = 15. You want to know this formula could be gotten though?
      (2 votes)
  • aqualine seed style avatar for user Cory Kilpatrick
    By looking at the comments, I doubt I will get a response to my question, but I just want to confirm something. The last partial sum Sal did, he plugged in 6 in a_n and got 33/220. I used the sum function on my calculator, plugged in a_n and calculated it from 1 to 6 and got .388601. It's bugging me how we got two different answers. If I plug in the equation for a_n using x's in a summation formula, shouldn't we get the same answer?
    (2 votes)
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    • leaf green style avatar for user kubleeka
      What you've done is compute S1, S2, ...S6, then add those values up. Sn gives the sum of the first n values of the sequence a. By computing it your way, you've added together:
      6 copies of a1
      5 copies of a2
      4 copies of a3
      3 copies of a4
      2 copies of a5
      1 copy of a6

      which is not the value we're seeking. We just need the sum of the first 6 terms of a, which, by definition, is S6.
      (2 votes)
  • duskpin ultimate style avatar for user arikaeli
    Can anyone help me with this question I had on a test and was unable to solve using this section?
    Given an infinite geometric series (it is a decreasing series, and all terms are positive):
    Each term (except for the first) is 2/5 of the sum of the two terms on either side of it (the term before it and the term after it).
    Find the ratio of the series.

    Thanks!
    (1 vote)
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    • leaf green style avatar for user kubleeka
      So for all n>1, a(n-1)>a(n)>a(n+1)>0, since the series is decreasing and positive.
      a(n)=(2/5)(a(n-1)+a(n+1)), given that property you mention
      For some r, a(n+1)=ra(n), since the series is geometric. Also, a(n)=ra(n-1)

      Let's take these last equations and plug them into the second, and we get
      a(n)=(2/5)(a(n)/r+ra(n))
      Divide through by a(n) (which is positive, and therefore nonzero) and we get
      1=(2/5)(1/r +r)
      5/2=1/r +r
      5r=2+2r²
      2r²-5r+2=0
      (2r-1)(r-2)=0

      So the ratio is either 2 or 1/2. But the series is decreasing and positive, and a ratio of 2 would give an increasing positive sequence.

      So the ratio must be 1/2.
      (4 votes)
  • leaf green style avatar for user Amandeep Singh
    At How does one come up with the formula for the partial sum for a series?
    (2 votes)
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Video transcript

- [Voiceover] Let's say that you have an infinite series, S, which is equal to the sum from n equals one, let me write that a little bit neater. n equals one to infinity of a sub n. This is all a little bit of review. We would say, well this is the same thing as a sub one, plus a sub two, plus a sub three, and we would just keep going on and on and on forever. Now what I want to introduce to you is the idea of a partial sum. This right over here is an infinite series. But we can define a partial sum, so if we say S sub six, this notation says, okay, if S is an infinite series, S sub six is the partial sum of the first six terms. So in this case, this is going to be we're not going to just keep going on forever, this is going to be a sub one, plus a sub two, plus a sub three, plus a sub four, plus a sub five, plus a sub six. And I can make this a little bit more tangible if you like. So let's say that S, the infinite series S, is equal to the sum from n equals one, to infinity of one over n squared. In this case it would be one over one squared, plus one over two squared, plus one over three squared, and we would just keep going on and on and on forever. But what would S sub -- I should do that in that same color. What would S -- I said I would change color, and I didn't. What would S sub three be equal to? The partial sum of the first three terms, and I encourage you to pause the video and try to work through it on your own. Well, it's just going to be the first term one, plus the second term, 1/4, plus the third term, 1/9, is going to be the sum of the first three terms, and we can figure that out, that's to see if you have a common denominator here, it's going to be 36. It's going to be 36/36, plus 9/36, plus 4/36, so this is going to be 49/36. 49/36. So the whole point of this video, is just to appreciate this idea of a partial sum. And what we'll see is, that you can actually express what a partial sum might be algebraically. So for example, for example, let's give ourselves a little bit more real estate here. Let's say, let's go back to just saying we have an infinite series, S, that is equal to the sum from n equals one to infinity of a sub n. And let's say we know the partial sum, S sub n, so the sum of the first n terms of this is equal to n squared minus three over n to the third plus four. So just as a bit of a reminder of what this is saying. S sub n... S sub n is the same thing as a sub one, plus a sub two, plus you keep going all the way to a sub n, and that's going to be equal to this business, n squared minus three over n to the third plus four. Now, given that, if someone were to walk up to you on the street and say, okay now that you know the notation for a partial sum, I have a little question to ask of you. If S is the infinite series, and I'm writing it in very general terms right over here, so S is the infinite series from n equals one to infinity of a sub n, and the partial sum, S sub n, is defined this way, so someone, they tell you these two things, and then they say find what the sum from n equals one to six of a sub n is, and I encourage you to pause the video and try to figure it out. Well, this is just going to be a sub one, plus a sub two, plus a sub three, plus a sub four, and when I say sub that just means subscript, plus a sub five, plus a sub six, well that's just the same thing as the partial sum, this is just the same thing as the partial sum of the first six terms for our infinite series. It's just going to be the partial sum S sub six. And we know how to algebraically evaluate what S sub six is. We can apply this formula that we were given. S sub six is equal to, well, everywhere we see an n, we replaced with a six, it's going to be six squared minus three over six to the third plus four, so what is this going to be? Six squared is 36 minus three, so that's 33, and six to the third, let's see, 36 times six, I always forget, my brain wants to say 216, but let me make sure that that's actually the case. Six times 30 is 180, plus 36, yes, it is 216, so I guess I have, inadvertently, by seeing six to the third so many times in my life, I have inadvertently memorized six to the third power, never a horrible thing to have that in your brain. So this is going to be 216 plus four, so 220. So, S sub six, or the sum of the first six terms of the series right over here, is 33/220, and we're done. And the whole point of this is just so you kind of appreciate, or really do appreciate this partial sum notation, and understand what it really means.