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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.

## Want to join the conversation?

• At the end, isn't the partial sum a1+a2+a3+a4+a5+a6? Why did he only find a6 and call it good?
• 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.
• At , isn't that the Riemann's zeta function?
• Yes, it is zeta(2), which converges to (pi)^2 divided by 6, many proofs of this, one using the Fourier series
• 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.
• 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.
• How are partial sums different from finite series?
• A partial sum describes part of a series. So you can have a finite series with 6 entries, but if you want to sum a part of it. say, the first three, it's called a partial sum.
Hope that helps
• 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)
• 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.
• Sal, at , can't I simplify 33/220 and right it as 3/20? Why would be wrong in doing so?
• There's nothing wrong with doing that. It's not always necessary to simplify unless the problem specifically says to, but you're welcome to simplify it if you like. Either answer is equally valid.
• 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?
• (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?
• 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?
• 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.
• 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)
• 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.