Topic 10.7: Alternating Series Test for Convergence

Same rods, a different answer

This is the Salar de Uyuni in Bolivia, the biggest salt flat on Earth. After a rain, a thin sheet of water turns huge stretches of it into a mirror, and the sky seems to carry on right under your feet. We’ve laid a rail across it, with two racks of glowing rods beside it. The blue rods are one, a third, a fifth and so on, and the amber ones are a half, a quarter, a sixth, forever.

Every blue rod pushes the shuttle forward by its length, and every amber rod pulls it back. Taking them in turn, we’re adding one, minus a half, plus a third, minus a quarter, and on forever. Let’s speed it up. The moves get tinier and tinier, and the shuttle closes in on one spot, right around zero point six nine three. Now a second shuttle draws from the same two racks, but it takes one blue rod and then two amber rods, over and over.

It uses every rod exactly once, the same rods as the first shuttle. Now watch where it settles, at about zero point three four seven, exactly half as far. We added the very same numbers in a different order and got a different answer. That should bother you, because you’ve known since second grade that two plus three is three plus two.

By the end of this lesson, you’ll know which infinite sums you can trust. You’ll even be able to make this one add up to any number you want. But first we need to know why a sum like this settles at all, so let’s slow the first shuttle down and watch its first few moves.

Why it settles

The first rod is a whole unit long, so after one move the shuttle sits at one. That’s the first partial sum, S one. The half-unit rod pulls it back to one half, which is S two. Now watch the next move closely.

A third is shorter than the half it just went back, so the shuttle can’t make it all the way up to one. It stops at five sixths, in between. A quarter pulls it back, but a quarter is shorter than a third, so it stops at seven twelfths, above one half, trapped between the last two places it turned around.

Every move lands inside the bracket made by the two moves before it, and that bracket is as wide as the rod that made it. From above, the odd partial sums keep falling and the even ones keep rising. The rods shrink to zero, so the gap between them closes on a single point.

That point is the sum of the series, about zero point six nine three, which happens to be the natural log of two. Let’s try different rods. These go two, then back one and a half, then forward one and a third, then back one and a quarter, each a little shorter than the one before.

These rods alternate and keep getting shorter. What do you think the shuttle does in the long run? Now you fix these rods. This slider sets the number the rods shrink toward. Drag it until the two edges close, and watch what happens to the gap.

Shrinking isn’t enough on its own. The rods have to shrink all the way to zero, or the trap never closes. This trap should feel familiar. It works like the squeeze theorem from the very first unit. The odd sums are a ceiling coming down, the even sums are a floor coming up, and they squeeze the series onto one number.

Let’s write down what made that trap work. We call it the alternating series test, or the AST for short, and you can use it to justify convergence on the exam. It’s for a series whose signs alternate, so each term is plus or minus a positive number, b n, the length of the rod. The negative one to the n plus one is what flips the sign.

The first condition is that each rod is no longer than the one before it, at least from some point on. That’s what kept the shuttle from getting back past its last turning point. The second is that the rods shrink all the way to zero. That’s what closed the trap. If both are true, the series converges.

This test can never prove a series diverges. If the rods don’t shrink to zero, it’s the nth-term test that says the series diverges, and the alternating series test just stays quiet.

Let’s run the test on our first shuttle, one minus a half plus a third and so on. We’ll write it the way you would on the exam. The negative one to the n plus one makes the signs alternate, and the rods are b n equals one over n, always positive. One over n plus one is smaller than one over n, because the bottom got bigger, so the rods are decreasing.

The limit of one over n as n goes to infinity is zero, so both conditions hold. On the exam you’d finish with one sentence that names every fact and the test, like the terms alternate, one over n decreases, and its limit is zero, so the series converges by the alternating series test.

Now let’s do a twistier one together. It’s the series of negative one to the n, times n over n squared plus one. The rods are n over n squared plus one. The first few are a half, then zero point four, then zero point three, so they look like they’re shrinking.

But a few numbers can’t prove that every rod is shorter than the one before. How do we know it keeps going down forever? Treat the rod length as a function of x and take its derivative with the quotient rule. That’s x squared plus one, minus two x squared, on top, which simplifies to one minus x squared, over x squared plus one, squared.

Once x is past one, the top is negative and the bottom is positive, so the rods keep shrinking. On the exam, a derivative like this is a solid way to show the terms decrease. The bottom grows like n squared while the top only grows like n, so the rods shrink to zero, and the series converges by the alternating series test.

Now try one on your own. Look at the series of negative one to the n, times n over two n plus one, and decide what you can conclude.