Topic 10.4: Integral Test for Convergence

Books into rectangles

To prove the tower never settles, we’re going to stop thinking of each push as a length and start thinking of it as an area. Take the little step that each book sticks out past the one below it, and stand those steps up side by side as rectangles, every one of them exactly one unit wide.

Since every rectangle is one unit wide, its area is just its height, which is that book’s push. The total area of all the rectangles is exactly the tower’s reach. Add the curve y equals one over two x. It runs through the top left corner of every rectangle, since at x equals one it’s a half, at two it’s a quarter, at three it’s a sixth.

The curve is heading down, so every rectangle pokes up above it, and that means the rectangles together have more area than the region under the curve. With eight books the bars run from one to nine, so the reach is bigger than the area under the curve from one to nine. The antiderivative of one over two x is a half times the natural log of x, so that area is a half times the log of nine.

Let’s call the number of books capital N. With N books the bars run from one to N plus one, so the reach is at least a half times the natural log of N plus one, and that log never stops growing. Wherever you put a wall, that area eventually gets past it, and the reach is bigger still. That’s the proof, and in our ideal model it means no wall can ever stop this tower.

There’s a second way to line up these rectangles, and I’d like you to find it. Drag the row sideways until every rectangle except the first fits underneath the curve.

Now the rectangles cap the reach from above, and that cap is the leftover half plus half the log of N. That means the reach is trapped between two curves, and both of them keep climbing forever. The row you slid is the picture that proves a series converges, so let’s make that same move with the twin. Its pushes become rectangles under the curve one half to the x, already slid one unit left, so we can put a ceiling on their total.

The area under that curve, all the way out to infinity, is finite, about point seven two, so the twin’s reach can never pass a half plus point seven two. That’s a ceiling for the twin, a little above its real wall at one. Its total only grows and can never pass a ceiling, so it has to settle.

On the board, the rectangles that poke out above the curve show divergence. When the area is infinite, so is the sum. The ones tucked under the curve show convergence. When the area is finite, the sum is too. Either way, the series and the integral share the same fate.

Suppose a positive, decreasing curve has an area of exactly three from one to infinity, and its heights at the whole numbers are the terms of a series. Can that series run off to infinity?

It only works for the right kind of curve. The terms have to be positive, so the total only ever grows, the way our towers do. The curve has to be decreasing. Over each rectangle it’s then highest at the left end and lowest at the right, which is why one row covers the area and the slid row fits underneath. It also has to be continuous, so the area under it makes sense.

The integral test

Let’s turn our rectangle argument into a test you can use on other series. Suppose f is positive, continuous and decreasing for x at least one, and each term a sub n equals f of n. Then the series and the improper integral of f from one to infinity either both converge or both diverge.

Let’s prove the big one properly, the harmonic series, one plus a half plus a third and so on. Our tower’s reach is exactly half of it. Before we use the test, let’s check the conditions. One over x is positive and continuous for x at least one, and its derivative, negative one over x squared, is always negative, so it’s decreasing.

We can’t plug in infinity, so we write the integral as a limit and integrate to a finite b first. That gives the natural log of b minus the natural log of one, and as b runs off to infinity, the log of b grows without bound. The integral diverges, so the harmonic series diverges by the integral test, and so does our tower’s reach, which is half of it.

Let’s try the sum of one over n squared. I’ll give you a head start. The integral of one over x squared from one to infinity comes out to exactly one, so what do you think that tells you about the series?

Now check it properly. One over x squared is positive, continuous and decreasing for x at least one. The integral from one to b is one minus one over b, and that heads to one as b goes to infinity. The integral converges, so the series converges by the integral test.

The series doesn’t add up to one, though. The first two terms alone, one plus a quarter, already make one and a quarter, so the test tells you whether a series settles, not where. The picture still pins it down, because the total is at most the first term plus the area, so it lands somewhere between one and a quarter and two. The exact answer is pi squared over six, a number mathematicians hunted for decades until Euler found it.

Now I’d like you to try one, the sum of n times e to the negative n squared. Work out the improper integral of x times e to the negative x squared from one to infinity, to three decimal places.

Now let’s write it up for a grader. A full integral test answer has three pieces, and graders check for each one. First come the conditions, and they’re about the function, not the series. f of x is positive and continuous, and by the product rule its derivative is e to the negative x squared times one minus two x squared. One minus two x squared is negative once x is at least one, so f is decreasing.

Second is the improper integral written as a limit, with b going to infinity in every line. Plugging in infinity, like writing e to the negative infinity, isn’t a limit calculation, and graders don’t accept it. Third is the value and the conclusion. Since the integral converges to one over two e, the series converges by the integral test. Name the test, and make the sentence about the series.

Start the integral where the series starts, as long as the conditions hold from there, so a series from n equals zero gets an integral from zero. Some functions climb before they fall, like the natural log of x over x, which rises until x equals e and only then starts decreasing. You just start the test at three, because the first two terms can’t change whether a series converges. Now check the conditions for these three series. Which one still fails them, however far along the series you start?

A meter over the sea

Let’s go back to the cliff and keep that promise. That pink line is one full meter out over the sea, which works out to twenty-five sixths book lengths, about four point one seven. Our proof says the reach is always at least a half times the log of N plus one. Once that lower bound reaches twenty-five sixths, we know the tower reaches at least that far.

Use the calculator to find the smallest number of books that this bound guarantees will reach the one meter line. Let’s test it. Pour in books until the tip passes the one meter line, then use the minus one and plus one buttons to find the last count that still falls short. That’s a tower about seventy meters tall, as high as a twenty-story building, with its tip hanging a full meter out over the water.

Now pick your own wall. Drag the pink line anywhere you like, out to about a meter and a third, and then add the number of books that our bound says is enough to get you there.

Drag that wall out to a hundred meters, though, and our two bounds say you’d need a number of books with three hundred sixty-two digits. The tower diverges, but it takes its sweet time. If the terms of a series don’t go to zero, it diverges by the nth-term test, and if they do, that test can’t tell you anything.

When the terms come from a positive, continuous, decreasing function, the series and its improper integral share a fate, though the integral tells you whether the series settles, never the number it settles on. Our tower’s pushes shrink to nothing, and it can still reach past any line you draw over that sea, and now you can prove it.