DinS Written on 2018/4/28

As I said before, convergence is a major topic in the field of series. Actually if a series doesn’t converge, we don’t have much to do with it. Here we’ll introduce commonly used convergence tests.

1. Comparison test

This is the most basic convergence test.

Statement:

Proof:

Since a_k >= 0, we have S_n <= S_n + a_n+1 = S_n+1. This means that S_n is monotone.

Since a_k <= b_k, we have . This means S_n is bounded.

Now that S_n is a bounded monotonic sequence, by monotone convergence theorem we know the limit of S_n exists. In other words, converges.

Let’s see an example.

Solution:

This is clear because the first term is 1. If we can prove all terms behind 1 converge, then adding 1 also converge. We do this because we’ll be using telescoping later.

We have a fact here.

This fact holds because n² – n < n². Notice that it’s because n starts from 2 we don’t have to deal with the case when denominator is 0.

By comparison test we have:

Now let’s prove converges.

It’s a telescoping series. . So we have:

Done.

2. Ratio test

This is an easy-to-use convergence test.

Statement:

Proof:

The most valuable condition we have is the limit one. It can provide some insight. What does it actually say? Well, it says that if n is big enough, a_n+1 = L*a_n. This means there’s a ratio between each term. This should remind you of geometric series. In fact we’ll use that later.

Now let’s get start. Pick a small number ε, which is a constant. Let r = L + ε.

Once we’re done picking ε, we can also find a constant N. As long as n >= N, .

What does that mean? Without the ε, if n is big enough the ratio approaches to L. Now we’ve picked a small number and set the original ratio a little bigger. The consequence is that if n is big enough the original ratio will be smaller than L + ε. What about the N? Only if n is big enough the above statements holds true. That means we can always find a constant N, so that whenever n >= N, n crosses the line of “big enough”.

Now that we’ve the N, we can split the original series into two parts.

The first part, from a₁ to a_N-1, though numerous, adds up to a constant, because N-1 is a constant. There’re only finite terms and each term can be worked out.

The second part, from a_N to infinity, is hard to deal but remember we have the ratio. We can get

This looks exactly like a geometric series, if we overestimate each term a bit. Let’s do that, and the second part will look like this:

This is geometric series! The good thing is we’ve already studied it well enough.

If ratio r < 1, the series converges. Besides, we’ve overestimated the series. Thus we have

By comparison test also converges.

You can work out the divergence case similarly.

OK, the last mystery. The key here is r, but we’re dealing with limit L. Since ε is small enough, we can get rid of it without a problem. We can just use L to judge whether the series converges or diverges.

After all the long proof, let’s reap the fruit.

By ratio test we have

Since the ratio is less than 1, the series converges.

Pretty simple.

One caution: The ratio is not the limit of the series.

3. Root test

Root test is similar to ratio test.

Statement:

In general ratio test is easier to use than root test.

4. Integral test

This is a very meaningful test because it offers some insight to the relationship between series and integral. We should remember that no math fact is an island.

We know that we can express a sequence in terms of a function.

If we draw it on the coordinate we get something like this.

Since n is a natural number, the domain of the function is 1, 2, 3, 4…

Thus the width of each rectangle is 1. In other words, the area of each rectangle is 1*y = f(n) = a_n.

If we add all rectangles up we have . That’s series.

Now look at the function side. We can do integral to calculate the area under the curve.

Notice that the curve is under the rectangle. This means that we have

The easy conclusion is that

Because this means the series is not bounded.

Once you understand the relationship between series and integral, see a formal statement.

Statement:

Suppose f is decreasing and positive, and a_n = f(n).

Quite simple and to the point.

Proof:

We’ve already proved the case when series diverges. Let’s now look at the other case.

For a constant N, we have

Seems complicated. Let me explain this a bit.

The first inequality holds because f is, as supposed, positive. When you add less terms up, the sum should be less or equal.

The second inequality holds because f is, as supposed, decreasing. That ceiling notation means we take the next bigger number than x. for example, if x = 3, after applying ceiling we take 4 instead. The messy part is the sum and integral. Well, we’re just rewriting the integral to a sum. Look at the graph above. stands for area for each rectangle. The sum notation says we’re adding them all up. Since f is decreasing, the sum of all rectangles will surely left certain curved areas behind. That’s why the sum of rectangles is less than the integral directly.

If the above inequality gets you, proceed next.

Applying ceiling we make x to n+1 because x and n are corresponding.

is just the area of rectangle. By definition we’ve already showed that the area of rectangle is the term in sequence. As a result we replace integral with a_n+1.