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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)?
• 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.
• 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
• 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?
• 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.
• I cant get one point why "if |r|<1, it converges"? It's at 2.18. If you explain, i will appreciate.
• 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.
• 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.