AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 10Lesson 1: Defining convergent and divergent infinite series
- Convergent and divergent sequences
- Worked example: sequence convergence/divergence
- Sequence convergence/divergence
- Partial sums intro
- Partial sums: formula for nth term from partial sum
- Partial sums: term value from partial sum
- Partial sums intro
- Infinite series as limit of partial sums
- Partial sums & series
Partial sums: formula for nth term from partial sum
The partial sum of a sequence gives as the sum of the first n terms in the sequence. If we know the formula for the partial sums of a sequence, we can find a formula for the nth term in the sequence.
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- I have a problem with this.
If you put n=1 into the S(n) formula, you get that the sum of the first 1 terms = 2/11.
Now if you look at his a(n) formula that he works out and put n=1 into it, it does not equal 2/11. It equals 9/110
So the sum of the first 1 terms is 2/11, but the first term is not 2/11. Is there something I don't understand? Thanks in advance.
EDIT: OK I watched again and heard the part where he says his a(n) formula is only true for n>1. Why is that?(24 votes)
- Let n = 1, then a(1) = S(1) - S(0) and S(n) = (n+1)/(n+10) that implies S(1) = 2/11,S(0) = 1/10 but S(0) = Sum of first 0 terms which is equal to zero ( S(0) = 0 ) and that is a contradiction. So the formula a(n) = S(n) -S(n-1) works only for n > 1. For n = 1, a(n) = S(n) and that make sense because a(1) is first term and S(1) is sum of first 1 term. Hope this is clear to you.(33 votes)
- How do you do the reverse of this process? How do you get the equation for Sn out of the equation for An?(12 votes)
- This process is not reversible, because a(1) is unknown(s(1) doesn't equal to a(1)).
However, given a(n), that means you know all the terms in the series, just sum a(1)...a(n) and you will get s(n), e.g: the summation of an arithmetic series is (a(1)+a(n)/2)*n(4 votes)
- Hi! I don't understand why at00:20all of a sudden the sum goes from a1 + a2 + .... + a(n-1) + an... Why the a(n-1)? I don't get it!(5 votes)
- Hi Emma,
The a(n-1) is the term right before the last term. e.g.:
If we have a sum of terms, it would look something like this;
a1 (first term) + a2 (second term) + .... (all in between) + a(n-1) (term right before last or nth term) + an (last term/nth term).
Hope this helps,
- Convenient Colleague(5 votes)
- So.... I get that S(n)-S(n-1) gives you the rule for a(n), but can we use the same logic and do S(n+1)-S(n) instead? Thanks!(6 votes)
- Think about it. S(n+1) - S(n) gives you, "a sub n + 1", not "a sub n".(4 votes)
- Ok I really hope Sal goes into how this is analogous to derivatives and integrals of continuous functions. Does anyone know any place that goes in depth into that connection?
The partial sum of the infinite series Sn is analogous to the definite integral of some function. The infinite sequence a(n) is that function. Therefore, Sn can be thought of as the anti-derivative of a(n), and a(n) can be thought of like the derivative of Sn.
Sooo, trying that out, I took the derivative of Sn using the quotient rule to try to find a(n)
And that's where I think you find the fundamental difference between the discrete and continuous realms that I really want Sal to dig into.
The derivative of S(n) is: 9 / n^2 + 20n + 100
The "actual" a(n) in the video: 9 / n^2 + 19n + 90
Only SLIGHTLY different, and that difference disappears as n gets large.
So the 2 questions are, why does the derivative of Sn not give you a(n), and why is it so close?(5 votes)
- We first graph S(n). We focus on n=x. To find a(x), we use S(x) and S(x-1) to make a right triangle, with the points as vertices, and legs of height a(x) and a base of 1.
Now, we look at the derivative. We can visualize the derivative as the limit of the slope of the line segment with endpoints (n,S(n)) and (x,S(x)) as n approaches x. One of these lines will have
(x-1,S(x-1)) as an endpoint, which we use our previous right triangle to find its slope being a(x). This means a(x) is an approximation of the derivative of S(n) at n=x.
This is why they aren't exactly the same- but more like approximations, but at the same time similar. I hope this helps.(2 votes)
- Why is it for n>1 why wouldn't n = .5 work with that formula(2 votes)
- Because for series you are takign sequential elements. Like things in a line and adding them up. When dealing with things (or people) in a line there aren't any people in between two others, by which I mean if you put someone between the first and second person already there, then it would change so the new person is the second person, and the person who was second is now third. I really hope that made sense.
Also, similarly, it cannot be less than 1 at all because series deal with ordered elements. So the first element where n=1 HAS to be the first.
Sometimes you can create a function that models a series that can be negative or in between two numbers, but a series itself deals only with terms where n is an integer.(1 vote)
- I'm getting the right answers, but in double checking and making sure I understand what's going on, things aren't adding up. Here's what I'm thinking:
S(n) = (n+1)/(n+10) and a(n) = 9/(n+9)(n+10);
S(3 ) = (3+1)/(3+10) = 4/13 = 0.30769231 = (a(1)+a(2)+a(3)) = (9/(1+9)(1+10))+(9/(2+9)(2+10))+ (9/(3+9)(3+10)) = (9/110)+(9/132)+(9/156) = 0.20769231
Which is not right. There seems to be a difference of a(0) i.e. 9/(0+9)(0+10) = (9/90) = (1/10) = 0.1
Where am I going wrong?
EDIT: OK, I got it. And I see now that my mistake is essentially the same as Sean's below. S(3) actually equals (S(1)+a(2)+a(3)) not (a(1)+a(2)+a(3)) in the terms by which I was thinking about it.(2 votes)
- I have a problem.
Show that S(n) of this is 25/48.(1 vote)
- First of all, the arbitrary term should be 1/n·(n+4), not 1/n·(n+1).
But okay, let's try to find the sum from n=1 to ∞ of 1/n·(n+4).
We'll start by rewriting this with partial fractions. So we need to find A, B such that 1/n·(n+4)=A/n+B/(n+4).
Since this holds for all n, we can plug in n=-1 and n=1 to get two equations:
Clean up the fractions to get -1=-3A+B and 1=5A+B.
Subtract the second equation from the first to get -2=-8A, or A=1/4. So B=-1/4.
So now, we're seeking the sum from n=1 to ∞ of (1/(4n)-1/4(n+4)).
Write out the first few terms:
And now, notice that 1/20 is both added and subtracted. In fact, almost every term will be added and subtracted, leaving us with
- why for n greater than one ?(1 vote)
- a(1) = S(1) - S(0), but this won't work because the sum of the first zero terms must be 0, so n has to be greater than 1.(2 votes)
- Us there a shorter way to solve partial sums when the starting index is greater than one?(1 vote)
- [Narrator] Nth partial sum of the series, we're going from one to infinity, summing it a sub n is given by. And they tell us of the formula for the sum of the first n terms. And they say write a rule for what the actual Nth term is going to be. Now to help us with this, let me just create a little visualization here. So if I have a sub one plus, plus a sub two, plus a sub three, and I keep adding all the way to a sub n minus one plus a sub n. This whole thing, this whole thing that I just wrote out. That is s sub n. This whole thing is s, let me this whole thing is sub n, which is equal to n plus one, over n plus 10. Now if I wanna figure out a sub n, which is the goal of this exercise, well I could subtract out the sum of the first n minus one terms. So I could subtract out this. So that is s, that is s sub n minus one, and what would that be equal to. Well wherever we see an n, we'd replace with an n minus one. So it'd be n minus one plus one over n minus one plus 10, which is equal to n over n plus nine. So if you subtract the red stuff from the blue stuff, all that you're gonna be left with is the thing that we're gonna solve for. You're gonna be left with a sub n. So we could write down a sub n is equal to s sub n, is equal to s sub n minus s sub n minus one S sub n minus one. Or we could write, that is equal to this stuff. So this is the n plus one over n plus 10 minus, minus n over n plus nine. And this by itself, this is a rule for a sub n. But we could combine these terms, add these two fractions together. And this is actually going to be the case for n greater than one. For n equals one, s sub one is going to be, well you can just, a sub one is going to be equal to s sub one. But then for any other n, we could use this right over here. And if we want to simplify this, well we can add these two fractions. We can add these two fractions by having a common denominator. So let's see, if we multiply the numerator and denominator here times n plus nine, we are going to get, this is equal to n plus one times n plus nine, over n plus 10 times n plus nine. And from that, we are going to subtract. Let's multiply the numerator and denominator here by n plus 10. So we have n times n plus 10, over n plus nine times n plus 10. N plus nine times n plus 10. And what does that give us? So, if we simplify up here, we're gonna have, this is n squared plus 10 n plus nine, that's that. And then this right over here this is n squared plus 10 n, within that red color. So this is n squared plus 10 n and remember we're gonna subtract this. And so, and we are close to deserving a drum roll, a sub n is going to be equal to our denominator right over here is n plus nine times n plus 10. And we're gonna subtract the red stuff from the blue stuff. So you subtract an n squared from an n squared and those cancel out. Subtract a 10 n from a 10 n, those cancel out. And you're just left with that blue nine. So there you have it, we've expressed, we've written a rule for a sub n, for n greater than one.