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## AP®︎/College Calculus BC

### Unit 10: Lesson 2

Working with geometric series- Worked example: convergent geometric series
- Worked example: divergent geometric series
- Infinite geometric series
- Infinite geometric series word problem: bouncing ball
- Infinite geometric series word problem: repeating decimal
- Proof of infinite geometric series formula

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# Worked example: divergent geometric series

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.3 (EK)

, LIM‑7.A.4 (EK)

Sal evaluates the infinite geometric series -0.5+1.5-4.5+... Because the common ratio's absolute value is greater than 1, the series doesn't converge.

## Want to join the conversation?

- Is there any way for us to know which infinity this series approaches (positive or negative)?(3 votes)
- You don't. Since the numbers alternate between positive and negative, at some point, you will add a very large negative number to an even larger positive number to an even larger negative number. Because of this alternation between positive and negative for the partial sum, it cannot be determined whether it approaches an extremely large negative value or an extremely large positive value. If the common ratio were positive and greater than 1, though, it would approach positive infinity, since you are just adding up larger and larger positive numbers.(4 votes)

- for the geometric series a+ar+ar^2+...

the sum of the first two terms is 24 and the sum to infinity is 27.

show that r=+-1/3. could you help, please(2 votes)- a/(1-r) = 27

a + ar = 24

You need to solve these two equations(1 vote)

- does diverge mean you cannot add up the sum?(2 votes)
- Basically, divergence means that the output value does not exist (could be positive or negative infinity, or does not approach a single non-imaginary value) as the limit approaches infinity.

In the case with summation, adding an infinite number of terms will result in divergence when their sum is positive or negative infinite or does not approach a finite value.

Conversely, convergence means that the output value does indeed exist (only one finite value). The sum of an infinite number of terms is convergent when adding an infinite terms approaches a single value.(3 votes)

- I cant get one point why "if |r|<1, it converges"? It's at 2.18. If you explain, i will appreciate.(3 votes)
- This comes with the proof of the geometric series, which i will cover here.

Imagine a series S going from 0 to a natural number n, inside of it is a constant a, multiplied by another constant r to the power of n, this means that we can write it as the following: a + ar^2 + ar^3 + ... ar^n-1, now if we multiply the series by r, we get ar + ar^2 + ar^3 + ... ar^n, now if we subtract S and rS we get: S - rS = S(1-r) = a + ar - ar + ar^2 - ar^2 .... ar^n-1 - ar^n-1 - ar^n, notice how all terms cancel except a and ar^n, so we get the following:

S(1-r) = a - ar^n, divide both sides by 1-r we get:

S = (a/1-r) - (ar^n/1-r), now if n is infinite, there are two possibilities, either r is less than 1 in which case the second term approaches zero and the infinite series converges, or r is more than 1 in which case, the 2nd term approaches infinity and the series diverges, i hope that this was clear.(1 vote)

- hello, I noticed that the geometric series started from n=0. Is there a reason why it doesn't start at 1? Personally, I think it might just be a matter of preference but I may be wrong. Thank you(1 vote)
- n is the number of terms.

Having the power of series from 0 to n-1 mean there are n terms because:

(n-1) - 0 + 1 = n

Think of the following question:

How many number are there from 0 to 150:

Answer: 150 - 0 + 1 = 151

Since we need to include 0 we have plus 1. Arguably you could say why not have the powers from 1 to n. In this case the first term is ar, however in my experience things become a bit confusing.

E.g. for the sequence 2, 4, 8 etc this will cause a = 1 instead of 2 since r = 2 => ar = 2a = 2 => a=1 and .

So I would recommend the power of a geometric series are defined from 0 to n-1.(2 votes)

- what does it mean to say a series is bounded or is not bounded(1 vote)
- If you have a series F, and have another series L; then if series F is
**always**less than series L for**all**terms n, then series F is bounded by series L because it can't ever be greater than it. This is useful for proving divergence or convergence in series.

Hope this helps,

- Convenient Colleague(1 vote)

- Does the fact that the series does not converge necessarily mean that the series diverges or does it just mean that the test is inconclusive?(1 vote)
- In this case, it means that the series diverges. This test is always conclusive -- if |r| < 1, it definitely converges, and if not, it definitely diverges.(1 vote)

- what happens when n=1?(1 vote)
- The n=0 under the sigma sign just indicates that for the first term in this series, you plug in 0 for n in the formula:

(-0.5)(-3)^0

Because (-3) to the zero power is just 1, your first term is (-0.5)(1) or -0.5.

If n=1, the first term in the series would have to be when you plug in 1 for n in the formula:

(-0.5)(-3)^1

Which gives us 1.5.

This is of course the second term in the first series, where we were given n=0. N is just the starting value, and tells us what the first term of the series is going to be. In this case, the series are the same, and the value of n just tells us which term we start with. For n=0, it is -0.5, and for n=1, it is 1.5.

I hope this answered your question.(1 vote)

- I don't really get this methodology of divergent and convergence thing? I just don't get the concept of what and why we are calculating a limit, and what does it even mean to "Converge" on 12, huh?(0 votes)
- As you add more and more terms to your sum, each term can either be big enough that it makes the sum approach infinity as the number of terms increases, or it can be small enough that it makes the sum approach a specific number (not infinity) as the number of terms increases. A divergent sum is one that gets bigger and bigger as you add more terms, and a convergent sum is one that "converges" or gets closer and closer to a number (if it converges on 12, it just means that if you add an infinite number of infinitely tiny terms in the series, it all adds up to 12). For a sum to be convergent, it both has to approach zero as the number of terms approaches infinity (which is why you use a limit to check if this is true), and it also has to pass some different tests we can use to analyze series.(2 votes)

## Video transcript

- [Voiceover] So we've got
this infinite series here, and let's see, it looks
like a geometric series. We're gonna go from this
term to the second term, we are multiplying by negative three, and then to go to the next
term we're gonna multiply by negative three again. So it looks like we have a
common ratio of negative three. So we could actually rewrite this series as being equal to negative 0.5, I could say times negative
three to the zero power, negative three to the zero power, plus negative zero, or maybe I can just keep writing this way, minus
0.5 times negative three to the first power, times
negative three to the first power, minus 0.5, minus 0.5 times negative three to the second power, negative three to the second power. And we're just gonna keep going like that. And we could just say we're
just gonna keep having minus 0.5 times negative three to each or to a higher and
higher and higher powers. Or we could write this in sigma notation. This is equal to the same thing, as the sum from, let's say
n equals zero to infinity, we're just gonna keep
going on and on forever. And it's gonna be this
first, it's gonna be, you can kind of think the
thing we're multiplying by negative three to some power, so it's gonna be negative 0.5, actually, I'm just gonna
do that yellow color. So it's gonna be negative 0.5 times negative three, negative, I'm gonna do
that blue color, so times negative three to the nth power. Here this is when n is
zero, here's n is one, here is n is equal to two. So we've been able to rewrite
this in different ways, but let's actually see
if we can evaluate this. And so we have a common
ratio of negative three. So our r here is negative three. And the first thing that
you should think about is well, in order for this to converge, our common ratio, the
magnitude of the common ratio, or the absolute value of the common ratio needs to be less than one for convergence. And what is the absolute
value of negative three? Well, the absolute value of
negative three is equal to three which is definitely not less than one. So this thing will not converge. This thing will not converge. And even if you look at
this, it makes sense, 'cause the magnitudes
of each of these terms are getting larger and larger and larger. We're flipping between
adding and subtracting. But we're adding and
subtracting larger and larger and larger and larger and larger values. Intuitively, when things
converge you're kind of, each successive term tends
to get diminishingly small or maybe it cancels out in some type of an interesting way. But because the absolute
value of the common ratio is greater than or equal to
one in this situation, this is not going to converge to a value.