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## Integral Calculus

### Course: Integral Calculus > Unit 5

Lesson 1: 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

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# Infinite series as limit of partial sums

Infinite series are defined as the limit of the infinite sequence of partial sums. Since we already know how to work with limits of sequences, this definition is really useful.

## Want to join the conversation?

- How do you come up with the formula for partial sums of non geometric series? The limit strategy is good to evaluate the sum but it all seems kind of useless if you cannot come up with the formula.(25 votes)
- There are several ways. One is to use algebra to deduce a formula, like Gauss did with S = 1,2,3,4,...(http://mathandmultimedia.com/2010/09/15/sum-first-n-positive-integers/). Another is to use established partial sums to derive new ones, like a generalization of Gauss' method to any arithmetic series. A third way is to use induction. For example, given the series 1,3,5, ..., the partial sum looks a lot like n^2 as one does S1, S2, etc. You can then prove this inductively. First the series is 2n+1. We want to prove that n^2 = S_n, so plugging in n we see that n^2=n^2, therefore the next partial sum is the next term(2n+1) + the sum of the pervious n terms (n^2). Plugging in the next n into our partial sum formula we see that (n+1)^2 = n^+2n+1, which is what we got earlier. This shows that given a partial sum = n^2, all partial sums after that follows that pattern. Then we simply do 1+3 = 2^2 to prove that there is a partial sum = n^2. I imagine that there are more ways but those are the ways I can think of. I feel like on AP tests they will be given though.(1 vote)

- In this video the teacher refers to a term and its subscript (S sub 3 for example) as the sum of the first 3 terms, but in the next video each term and its subscript are evaluated separately for a value. Which is true or is one true sometimes and not other times?(7 votes)
- Depends on what
`S`

means. If you are talking about sequences and`S`

represents a sequence, then`Sᵢ`

will refer to the value of the i-th element in the sequence.

If you are talking about series, and`S`

represents the partial sum of the series, then`Sᵢ`

represents the partial sum up to the i-th item of the series.

There is nothing special about the letter`S`

, you can use any symbol you want to represent either sequences or series.(10 votes)

- What are the potential errors that arise if, when testing for convergence, we take the limit of the general formula of the last element in the sequence instead of the partial sum?(4 votes)
- Good question! Let's think about this.

Let 𝑎(𝑛) denote the 𝑛th term in a series and let 𝑆(𝑛) denote the 𝑛th partial sum. we know that if lim(𝑛 → ∞) 𝑆(𝑛) converges, then the infinite sum exists and is equal to that limit. However, what if lim(𝑛 → ∞) 𝑎(𝑛) converges? What does this say about 𝑆(𝑛)? Well we know that:

𝑆(𝑛 + 1) = 𝑎(𝑛) + 𝑆(𝑛)

We can take the limit of both sides:

lim(𝑛 → ∞) 𝑆(𝑛 + 1) = lim(𝑛 → ∞) [𝑎(𝑛) + 𝑆(𝑛)]

Suppose that the infinite series exists at 𝑘 (i.e. lim(𝑛 → ∞) 𝑆(𝑛) = 𝑘). Then we have:

𝑘 = 𝑘 + lim(𝑛 → ∞) 𝑎(𝑛) ⟹ lim(𝑛 → ∞) 𝑎(𝑛) = 0

So the sum can only converge if the terms are approaching 0. This makes sense intuitively because the terms need to get smaller and smaller if we have any hope of our sum "jumping" around less and less and "settling" on a single value. If the terms converged to any nonzero number, then naturally, the sum diverges as the sum would keep going on and on.

But it should be noted that this is a*necessary*but not a*sufficient*condition for series convergence. It is possible for the terms of a series to converge to 0 but have the series diverge anyway. The classic example of this is the harmonic series:

𝚺(𝑛 = 1) ^ ∞ [1/𝑛]

Obviously here, the terms approach 0, (lim(𝑛 → ∞) 1/𝑛 = 0) but in fact, this sum diverges! So the fact that the terms of a series approach 0 is a necessary but insufficient condition for series convergence. On the other hand, the fact that the partial sums of a series converge*is*in fact a sufficient condition for convergence because this is exactly what we define series convergence to be. An infinite sum exists iff the sequence of its partial sums converges.

Comment if you have questions!(16 votes)

- Shouldn't he have divided the n values by n^3 instead of n^2 because n^3 is the highest exponent?(3 votes)
- It has to be the highest exponent in the denominator.(2 votes)

- At3:25why does Sal divide, both the numerator and denominator by n^2?(2 votes)
- It is a method to more easily see what the end behavior will be. Hopefully you see that dividing the numerator and denominator by the same thing doesn't change the fraction.

Now, when you divide by the highest degree variable you get a bunch of terms with variables only in the denominators. When you look at a term that has a variable only in the denominator and its end behavior that term goes to 0, since one over a very large number is practically 0, and the larger ti gets the closer to 0 it gets.

This means only the terms that had degrees to match the largest degree you divided by are left. this means the function, whatever it is, becomes identical to a function with just those terms.

An example, if you look at (x^2 + x + 1)/(x^2 + 10000000x + 1000000000), if you compared the two at higher and higher x values they would look more and more like the same function. Specifically x^2 / x^2 = 1(2 votes)

- I was thinking about this when the tutor wavered on what to multiply by: if you multiplied by n^3/n^3, you would end up with a "convergence" at undefined. Further, if you multiplied by n^4/n^4, the sequence would be convergent! So by using one of three scalars -- all variations of 1 and therefore 100% equal -- you end up with three completely different answers?(1 vote)
- In all three cases, the infinite series actually diverges. When you multiply n^3/n^3, you end up getting 2/0, while on multiplying n^4/n^4, you get 0/0. Both values are undefined (in this case. 0/0 can be a defined value in some cases) and hence, the series diverges. Plus, you can graph this expression you get on multiplying n^3/n^3 and n^4/n^4 (as a matter of fact, whichever power of n^a/n^a you multiply, you get the same graph. Shouldn't be a surprise though) and you'll see that as n tends to infinity, the expression does too.(3 votes)

- heyy, what does s base infinity equals to?(2 votes)
- It means the partial sum of the first infinity terms aka the value of the infinite series.(1 vote)

- Aren't there are nested sums with infinite limits?(2 votes)
- If the limit of the partial sums is 0, does that mean that the sum of the series equals 0?(2 votes)
- Yes. because the sum of the series is defined as the limit of partial sum sub n with n approaching towards infinity. As far as I know (if I'm wrong please correct me), this can only happen if either a sub n oscillates around zero with positive and negative numbers of the same magnitude somewhere in the sequence (e.g. 5 - 7 - 3 + 7 - 5 + 3 + ... = 3 - 3 + 5 - 5 + 7 - 7 + ... = 0), or the partial sum always stays at zero as a sub n is also equal to zero.(1 vote)

- If the limit to infinity of the partial sum is a finite value, then is that finite value also the sum of the series?(1 vote)
- Yes. If the limit of the partial sums exists - is a finite value - then the series converges and the series equals the limit. Also see the answer below by sauj123, who answered with respect to the specific case of the limit being zero. Consider his reminder of the definition of an infinite series.(2 votes)

## Video transcript

- [Voiceover] Let's say that
we have an infinite series S so that's the sum from n = 1 to infinity of a sub n. We could write it out a sub 1 plus a sub 2 and we're just going to go on
and on and on for infinity. We're going to go on
and on and on forever. So, let's say, and I've written
it in very general terms let's say we have a formula
for the partial sums of S. We know that S sub n is equal to 2n to the third over n plus 1 times n plus 2. Now, my question to you is,
based on what I've just told you S is the sum in a very general way written this infinite series but
I have the partial sum. The sum of the first n terms of S is given by this formula right over here does this series converge or diverge? Does this thing converge
to some finite value or is it unbounded and does it diverge? Well, one way to think about this is the idea that our infinite series S is just the limit as n approaches infinity of our partial sums. So, what do we mean by that? Well, you could a sequence
of partial sums here. You have S sub 1, S sub 2, S sub 3 and you keep going so this would be the sum of the first term. This would be the sum
of the first 2 terms. This would be the sum of the first 3 terms and just think about what
happens to this sequence as n right over here approaches infinity because that's what this series is. It's the sum of the first, I guess you could say the first, infinite terms. It's the sum of all, you have an infinite number of terms here. Well, let's think about what this. The limit is n approaches
infinity of S sub n. That's just going to be the limit as n approaches infinity of this business right over here. 2n to the third power over n plus 1 times n plus 2 and there's several ways you could evaluate this. One way is you could just realize, "Hey, look in the bottom this is going "to be a second degree polynomial." On up here, you have a third degree so the numerator is gonna grow faster than the denominator so this
is going to be unbounded. So that will immediately tell you well this is gonna approach infinity so S is going to diverge but if you wanna do it a little bit less hand wavy than that we can actually do a little bit more algebra. Limit as n approaches infinity, 2n to the third power over, let's multiply this out, n squared plus 3n plus 2 and, let's see, we can
divide the numerator and the denominator by n squared. So, this is going to be the limit as n approaches infinity of, if we divide the numerator by n squared, you're going to have, actually, let's
divide the numerator and, well yeah let's
divide it by n squared so, if we divide the
numerator by n squared, we're gonna have 2n and
then the denominator divided by n squared you're gonna have 1 plus 3 over n plus 2 over n squared. Now, when you look at it like this, it becomes pretty clear this thing as n approaches infinity,
this thing is gonna towards infinity but this thing down here the denominator this
is gonna go towards 0. This is gonna go towards 0 so the denominator's gonna go towards 1. So, this whole thing, is the limit is gonna go to infinity
and since the limit of the partial sums goes to infinity that mean that this infinite series is not going to be a finite value. It's just going to diverge. So, this character right over here is going to diverge. In order for it to have converged, this thing should have come, this limit should have been some finite value. So, hopefully, that makes sense. All we say is, "Look, infinite series, "we had a formula for the partial sum "of the first n terms
and then we said oh look "the series itself, the infinite series, "you could view it as a limit of, "as n approaches infinity,
of the partial sum "S sub n and we said hey,
that approach infinity "this thing is diverging."