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## Working with geometric series

Current time:0:00Total duration:3:14

# Worked example: divergent geometric series

AP Calc: LIM‑7 (EU), LIM‑7.A (LO), LIM‑7.A.3 (EK), LIM‑7.A.4 (EK)

## 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.