An example of an infinite series.

Damian has two identical cups, cup 1 and cup 2. Cup 1 is initially full of water, and cup 2 is initially empty. He begins a procedure in which he exchanges the water among the cups. For the first turn of the process, he pours half of the water in cup 1 into cup 2. Next, he pours half of the water in cup 2 back into cup 1, and then half of the total water in cup 1 back into cup 2 (see illustration below). He continues this process for some time.

If Damian were to continue this process indefinitely (for an infinite amount of turns), how much water would remain in each cup?

Set up

Cup 1 is initially full, so the total volume of water it holds is considered to be 1. A “turn” is each time Damian pours half of the water in one cup into the other. Let’s create a table to see how the volume of water changes after a certain number of turns.

It seems that the volume of water in cup 1 is approximately 1/3 for odd numbered turns and 2/3 for even numbered turns, and the volume in cup 2 is 2/3 for odd numbered turns and 1/3 for even numbered turns. This gives us a pretty good idea that if this process were continued indefinitely, the volume of water in each cup should oscillate from 1/3 and 2/3, but how can we actually prove this?

What to do in a nutshell

We’ll have to figure out a formula for the volume of water in each cup after some number of turns and take its limit as the number of turns approaches infinity.

Getting started

We’ll have to express the volume of water in each cup in a different manner. It looks more complicated, but a very interesting pattern becomes evident. Closely examine the way the volume in each cup is expressed on the chart below:

 Because the total volume for both cups must always be 1, we couldn't end up with more or less than what we started out with. After we've expressed the volume of water in one cup, we can simply subtract it from 1 in order to have an expression for the volume of water in the other cup.  In addition, the exponent in the denominator represents the number of times the volume of water in each cup has been halved.

Let us from now on focus on even numbered turns of cup 1. Consider the pattern seen in turn 6, an even turn. The first term that appears in the numerator (2^n-1 ) is raised to a power that is one less than the number of the turn (turn n), and this is true for the denominator as well. The exponent also lessens by 1 as we progress through the series. Upon analyzing this pattern, we can write volume of water in cup 1 for the nth even term as the following series V_n:

If we distribute the negatives and remove the parentheses, we get the following alternating series:

*First of all, let me point out why the quantity (1-1/2) is negative. Look at the chart above. If we examine the expression of the volume in cup 1 for even turns, we’ll notice that there are an odd number of negatives that get distributed, ultimately leaving the quantity (1-1/2) as negative. We can conclude that for even turns, this portion of the series will always be negative, and because V_n applies specifically to even turns of cup 1, its final term will also be negative; however, we will soon see that this portion of the series can be ignored regardless of its sign.

Since that’s out of the way, let’s go ahead and break up V_n into two parts:

Let’s go ahead and divide the terms of the series by 2^n-1:

This is where it gets exciting. Notice that the terms of the series share a fixed ratio of 1/4; it’s a geometric series! It was pretty evident earlier that this is a geometric series, but doesn't it look so much cleaner now? To make V_n even more compact, we will rewrite it using sigma notation. Below I've provided the formula for the nth term of a geometric sequence because we will use it to rewrite the sum. The term a₁ is the first term of the sequence and r is the common ratio

This looks significantly better. This takes us back to the quantity (1-1/2) that became 2^-n after dividing by 2^n-1. Remember that we are working with an infinite series. As n becomes very large, 2^-n approaches 0. That’s why this term of the series is ignored, and it doesn’t matter to us whether this quantity is positive or negative; it’s zero either way.

We are interested in calculating the sum of this geometric series, so we can apply the following formula:

Notice that the geometric series we are dealing with has a common ratio that has an absolute value less than 1, and when such a number is raised to the power of some positive number greater than 1, the output decreases. For example,

0.5²=0.25 and 0.5³ is even smaller. If we take the limit of S_n as n approaches infinity, we will get the following formula:

This is actually the standard formula for an infinite geometric series. Note that it is only valid if the absolute value of the common ratio is less than 1.

We are now ready to apply this formula to V_n. We know that a₁ is equal to 1 and r is equal to 1/4:

This is our final solution, and it seems that our conjecture from the very beginning is correct! The volume of water held by cup 1 after many even turns is 2/3. From this, we can conclude that the volume held by cup 2 must be 1/3 for even turns. Well, what about odd turns? Look at the chart below:

The volume of water held by each cup will oscillate from 1/3 and 2/3 over and over and over and over…