Worked example: convergent geometric series | Series | AP Calculus BC | Khan Academy

Let's get some practice taking sums of infinite geometric series.

So, we have one over here, and just to make sure that we're dealing with the geometric series, let's make sure we have a common ratio.

So, let's see: to go from the first term to the second term, we multiply by ( \frac{1}{3} ). Then, to go to the next term, we are going to multiply by ( \frac{1}{3} ) again, and we're going to keep doing that.

So, we can rewrite the series as ( 8 + 8 \times \frac{1}{3} + 8 \times \left(\frac{1}{3}\right)^2 + 8 \times \left(\frac{1}{3}\right)^3 + \ldots ). Each successive term we multiply by ( \frac{1}{3} ) again.

So, when you look at it this way, you're like, okay, we could write this in sigma notation. This is going to be equal to…

So, the first thing we wrote is equal to this, which is equal to the sum:

The sum can start at zero or at one, depending on how we'd like to do it.

We could say from ( k = 0 ) to infinity. This is an infinite series right here; we’re just going to keep on going forever. So, we have:

[
\sum_{k=0}^{\infty} 8 \times \left(\frac{1}{3}\right)^k
]

Let me just verify that this indeed works, and I always do this just as a reality check, and I encourage you to do the same.

So, when ( k = 0 ), that should be the first term right over here. You get ( 8 \times \left(\frac{1}{3}\right)^0 ), which is indeed ( 8 ).

When ( k = 1 ), that's going to be our second term here. That's going to be ( 8 \times \left(\frac{1}{3}\right)^1 ), which is what we have here.

And so, when ( k = 2 ), that is this term right over here. So, these are all describing the same thing.

Now that we've seen that we can write a geometric series in multiple ways, let's find the sum.

Well, we've seen before, and we proved it in other videos, if you have a sum from ( k = 0 ) to infinity and you have your first term ( a ) times ( r^k ), assuming this converges—so, assuming that the absolute value of your common ratio is less than one—this is what needs to be true for convergence.

This is going to be equal to:

[
\frac{a}{1 - r}
]

This is going to be equal to our first term, which is ( a ), over ( 1 - r ).

If this looks unfamiliar to you, I encourage you to watch the video where we derive the formula for the sum of an infinite geometric series.

But just applying that over here, we are going to get:

This is going to be equal to ( \frac{8}{1 - \frac{1}{3}} ).

We know this is going to converge because the absolute value of ( \frac{1}{3} ) is indeed less than one.

So this is all going to converge to:

[
\frac{8}{1 - \frac{1}{3}} = \frac{8}{\frac{2}{3}} = 8 \times \frac{3}{2} = 12
]

Let's see: this could become, divide ( 8 ) by ( 2 ); that becomes ( 4 ), and so this will become ( 12 ).