1 Minus 1 Equals What? (Fun with sums and limits)

By Alex Beal
August 5, 2011

Here’s a cool problem I came across when reviewing for one of my calculus exams:

You begin with the numbers 0 through 1. Every iteration removes 1/3 from the remaining segments. As stated above, the expression for the total amount removed after m iterations is:

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The first iteration removes 1/3. The second removes 1/3 from the remaining two segments. The remaining segments are 1/3 long, and 1/3 of that is 1/9. We do that once for each remaining segment, so 2/9 is removed. And so on.

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What if we iterate an infinite number of times? Then we have a convergent geometric series. Finding the sum is easy:

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So, what’s the total amount or length of numbers removed? 1. That’s the entire length. It seems like we’ve removed the entire segment. But wait a minute. If we look at the image of the original question above, we see that after every iteration, there are 2n+1 segments remaining. When n=0, there are 2 segments. When n=1, there are 4 segments, etc. As we remove more and more, 2n+1 approaches infinity. In other words, the limit does not exist, and the number of segments goes to infinity.

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So, even though we’ve seemingly removed the entire length, there are an infinite number of segments remaining.