Topic 10.8: Ratio Test for Convergence

Each rod a fraction of the last

Every series so far had sizes like one over n or one over n squared. Some series build their terms from powers and factorials instead, like two to the n, over n factorial. Remember, n factorial is one times two times three, all the way up to n, and zero factorial is one.

Starting at n equals zero, the rods come out as one, two, two, four thirds, two thirds, so at first they actually get longer. If we keep making rods by this rule and lay them all end to end, do you think the chain ever stops?

After the fourth rod, every new rod is at most half the one before it. Each one fits inside a ghost rod that halves every time. Those first few rods add only a fixed length. After them, every rod fits inside the halving chain, a geometric series with a finite length, so the real chain has to stop too. It stops at about seven point three nine.

That gauge is the idea behind the ratio test. Take the limit of the next term over the current term, in absolute value, and call it L. If L is less than one, the terms eventually shrink at least as fast as a geometric series, so the series converges absolutely.

If L is bigger than one, the terms eventually grow, so they can’t go to zero and the series diverges. That includes a ratio that runs off to infinity. If L is exactly one, the test tells you nothing. We’ll see why in a minute.

Let’s run it on the series of negative one to the n, times three to the n, over n factorial. Put the next term over the current one inside absolute value bars, so the signs disappear. Dividing by a fraction means multiplying by its flip.

Write n plus one factorial as n plus one, times n factorial, and three to the n plus one as three times three to the n. Then the n factorials cancel, and so do the three to the n’s.

That leaves three over n plus one, and its limit is zero. Zero is less than one, so the series converges absolutely by the ratio test. Now let’s do one together, n factorial over ten to the n. Its first terms shrink, a tenth, two hundredths, six thousandths.

But the ratio is n plus one over ten, which blows up to infinity, so this series diverges. Its terms start growing after the tenth one, and the first few hid that. Now try one yourself. Use the ratio test on n squared over three to the n, and find L.

There’s one place the ratio test goes blind, and it’s when L equals one. Here are three series where that happens. For the harmonic series, the ratio n over n plus one goes to one, and the harmonic series diverges. For one over n squared, the ratio also goes to one, but that series converges absolutely.

Our alternating harmonic series gives one again, and it converges conditionally. All three give L equals one and do three different things, so L equals one tells you nothing. You have to reach for another test.

Suppose the ratio test gives L equals one for n plus two over n cubed plus one. What can you conclude from the ratio test alone?

Steer the sum

Back on the mirror, let’s settle the mystery, starting with just the first six rods. However we shuffle them, they add up to thirty-seven sixtieths, because with finitely many numbers the order never matters. An infinite series never runs out of rods. The gauges show that the blue rack alone is infinitely long, and so is the amber one, since each adds up like half the harmonic series.

If the shuttle is short of a spot, there’s always enough blue to push it past. If it’s past, there’s always enough amber to pull it back. Each overshoot is at most the rod that just went in, and the rods shrink to zero.

A conditional series travels forever but still settles. For that, both racks have to be endless. If one ran out, the other would drag the shuttle away. Endless racks are what let you aim it at any number you like.

Before the rule does it for you, try it yourself. Lay the rods one at a time, blue or amber, using at least ten rods, and land the shuttle within three hundredths of the flag at one. Drop the flag anywhere on the rail, the farther from zero point six nine the better, and the shuttle will use every rod exactly once and land on your number.

Now swap in the one-over-n-squared rods, the series that converges absolutely, and look at the racks. All the blue adds up to about one point two three, and all the amber to about zero point four one.

Now look at those totals and think about it for a second. With these rods, can you steer the sum all the way to one and a half?

Absolute convergence comes with a guarantee. If a series converges absolutely, then any rearrangement of its terms, or any way of grouping them with brackets, adds up to the same sum. Grouping is the easy half. Put brackets around pairs, and you’re only looking at every other partial sum, S two, S four, S six, so the sum can’t change. Rearranging is the part that needs absolute convergence.

A conditionally convergent series makes no such promise. The mathematician Bernhard Riemann proved that you can rearrange one to add up to any number at all, which is what you just did with your flag. That even goes for the series for pi over four. It converges only conditionally, so if you shuffle it the wrong way, you won’t get pi over four at all.

Which sums can you trust?

When the exam asks whether a series converges absolutely, conditionally, or not at all, you can work through it in this order. If the terms don’t go to zero, it diverges by the nth-term test. If you see factorials or powers like three to the n, try the ratio test.

Otherwise, drop the signs and test the absolute values with a p-series or a comparison. If that converges, the series converges absolutely. If the absolute values diverge but the series alternates, the alternating series test can still give you convergence, and then it’s conditional.

At the start, two shuttles used the same rods and stopped in different places. The alternating harmonic series converges only conditionally, so its order is part of its answer. You can trust the sums that converge absolutely, because they land on the same number however you shuffle them, and the conditional ones you can steer anywhere you want.