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## Infinite geometric series

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# Infinite geometric series formula intuition

## Video transcript

What I want to do is
another "proofy-like" thing to think about the sum of
an infinite geometric series. And we'll use a
very similar idea to what we used to find the sum
of a finite geometric series. So let's say I have a geometric
series, an infinite geometric series. So we're going to
start at k equals 0, and we're never going to stop. We're it's going all
the way to infinity. So we're never going to
stop adding terms here. And it's going to be our first
term times our common ratio. Our common ratio
to the kth power. Actually let me do
k and that color. k equals 0 all the
way to infinity. And so let's just call
this thing right over here, let's call this s sub infinity. We're going all the way to
infinity right over here. And so this, if we
were to expand it out is going to be equal to a
times r to the 0-- actually let me just write it out
like that which is just a. a times r to the 0 power plus
a times r to the 1st power. r to the 1st power. Plus a times r to the 2nd power. r to the 2nd power. Plus-- and we could just
keep going on and on and on. I think you get
the general idea. Now just like when
we tried to derive a formula for the sum of
a finite geometric series we just said, well what
happens if you take the sum and if you were to
multiply every term by your common ratio. Every term by r. So let's do that. Let's imagine this sum. And we're going to
multiply every term by r. And the reason
why I said this is "proofy" is this is not always
clear-- It's a little bit, when you're multiplying
something times infinite terms or an infinite sum, at
least this will be at least give you the general idea. Or when you start thinking
about and infinity, sometimes I just think about
things a little bit deeper. So r times this infinite sum? Well that's going to
be equal to-- We're just going to multiply
every term here times r. So a r to the 0'th power times
r is going to be a times r. a times r to the 1st power. Multiply this one times r? You're going to get a
times r to the 2nd power. a times r to the 2nd power. I think you see
where this is going. Multiply this one times r? You're going to get plus a
times r to the 3rd power. And we would just keep on going. We'd just keep on going. So let me just show that. So plus dot dot dot. Now what happens if
we were to subtract this sum from this top sum? So on the left
hand side, we could express that as our sum s sub
infinity minus our common ratio times s sub infinity. Is going to be equal to-- So when you subtract you're
going to have a times r to the 0'th power, which is
really just the same thing as a. That's just going to be
a. a times r to the 0 is just a times 1 which is a. We'll write in that same color. Is equal to a. But every other
term, you're going to have a times r to
the 1st, but you're gonna subtract a
times r to the 1st. You have a times r to
the 2nd, but you're going to subtract a
times r to the 2nd. So every other term is
going to be subtracted away. And this happens all
the way to infinity. It never, never ends. So the only term
that you're left with is just that
first one, is just a. And so now we can actually
try to solve for our sum. If you factor out
the s sub infinity, you are left with 1 minus r. 1 minus r. s times s, our sum, times
1 minus r is equal to a. Divide both sides
by 1 minus r, and we get that our sum, the
thing that we cared about-- And once again, this is
kind of an amazing result. That we're taking the sum of
an infinite number of terms and under the
proper constraints, we are going to
get a finite value. So this is going to be
equal to a over 1 minus r. So once again,
it's kind of neat. If let's say I had the sum,
let's say we started with 5, and then each time we
were to multiply by 3/5. So 5 plus 3/5 times 5 is 3,
times 3/5 is going to be 9/5. 9/5, or I'll multiply by 9/5
again-- Oh, sorry not 9/5. My brain isn't working right! 5 times 3/5 is going
to be 3 times 3/5. Is going to be-- 3
times this is going to be 9/5-- actually
that was right. My brain is working right. Times 3/5 is going
to be 27 over 25. Times 3/5 is going to be 81/125. And we keep on going on
and on and on forever. And notice these
terms are starting to get smaller and
smaller and smaller. Well actually all
of them are getting smaller and smaller and smaller. We're multiplying
by 3/5 every time. We now know what the
sum is going to be. It's going to be our
first term-- it's going to be 5-- over 1
minus our common ratio. And our common ratio
in this case is 3/5. So this is going to be
equal to 5 over 2/5, which is the same
thing as 5 times 5/2 which is 25/2 which is equal
to 12 and 1/2, or 12.5. Once again, amazing result. I'm taking a sum of
infinite terms here, and I was able to
get a finite result. And once again, when
does this happen? Well, if our common ratio--
if the absolute value of our common ratio--
is less than 1, then these terms
are going to get smaller and smaller and smaller. And you'll even see here it
even works out mathematically in this denominator
that you are going to get a reasonable answer. And it makes sense
because these terms are getting smaller
and smaller and smaller that this thing will converge. Even if r is 0. If r is 0, we're still
not dealing strictly with a geometric series anymore,
but obviously if r was 0, then you're really only going
to have this-- well, even this first term is
kind of under debate depending on how you
define what 0 to 0 is. But if your first term
you just said would be a, then clearly you'd just be left
with a is the sum, and a over 1 minus 0 is still a. So this formula that we just
derived does hold up for that. It does start to break down if
r is equal to 1 or negative 1. If r is equal to 1 then
as you imagine here, you just have a plus
a plus a plus a, going on and on forever. If r is equal to negative 1
you just keep oscillating. a, minus a, plus a, minus a. And so the sum's value keeps
oscillating between two values. So in general this
infinite geometric series is going to converge
if the absolute value of your common ratio
is less than 1. Or another way of saying
that, if your common ratio is between 1 and negative 1.