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.

# 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.

## Want to join the conversation?

• 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? •  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.
• How do you do the reverse of this process? How do you get the equation for Sn out of the equation for An? • Hi! I don't understand why at all of a sudden the sum goes from a1 + a2 + .... + a(n-1) + an... Why the a(n-1)? I don't get it! • 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! • 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? • 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.
• Why is it for n>1 why wouldn't n = .5 work with that formula • 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. • I have a problem.
1/(1×5)+1/(2×6)+1/(3×7)+....+1/n(n+1)+.....
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:
1/-3=A/-1 +B/3
1/5=A/1+B/5
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:
1/4-1/20+1/8-1/24+1/12-1/28+1/16-1/32+1/20-1/36+...
And now, notice that 1/20 is both added and subtracted. In fact, almost every term will be added and subtracted, leaving us with
1/4+1/8+1/12+1/16=12/48+6/48+4/48+3/48=25/48.
QED
•  