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### Course: Integral Calculus>Unit 5

Lesson 1: Convergent and divergent infinite series

# Partial sums: formula for nth term from partial sum

This video shows how to find a rule for the nth term (aₙ) in a series. We start with the formula for the sum of the first n terms (sₙ). Then, we subtract the sum of the first n-1 terms (sₙ₋₁) from sₙ to get aₙ. Finally, we simplify the expression.

## 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?
• 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
• 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!
• 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
• 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!
• Think about it. S(n+1) - S(n) gives you, "a sub n + 1", not "a sub n".
• 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.
• What exactly is a "rule" in this context? I haven't encountered that term before
(1 vote)
• A rule is quite simply the general formula. You're basically asking "what general formula for a_n can I derive so that all values of n follow it and I can derive a_n for any value I want?"
• 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)
• In I don't understand how the given problem was solved. I kept getting different answers
• Are you having trouble combinding fractions?
(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