### Using area under a curve to prove or disprove convergence

The integral test is a very reliable way to establish the convergence of a series. The basic idea is that if the nth term of a series, an, has a functional form f(n), then the curve f(x) will overlie a bar graph of the series representing its terms. If we integrate f(x) between some fixed starting point, a, and infinity (an "improper integral"), and the integral is finite, then the sum, which is less than f(x) for all of x, must also be finite – that is, it converges.

Here's a picture of the situation:

Here's another way to say it: If the area under f(x) reaches some limit, then anything contained inside that area must also have a finite area.

Symbolically, if our sum has terms an, like this:

$$S = \sum_{n = 1}^{\infty} \, a_n$$

and the an can be generated by a function, f(n), then the improper integral (what a terrible name – they're quite "proper") of the function from any point in the series (I've chosen n = 1 here) to infinity is:

$$\int_1^{\infty} f(x) \, dx = \lim_{R\to\infty} \int_1^R f(x) \, dx$$

If this integral has a limit, the series must converge because it is trapped within a finite area – an area guaranteed never to grow past a certain limiting size.

On the other hand, some areas continue to grow (albeit more slowly) with increasing x. If each term of a series is under such a function, then the sum of the series can also grow without bound.

### When the integral test doesn't work

Remember a very important feature of calculus: While every function we can write down has an analytical derivative, that is, one we can do with pencil and paper, not every function we can write down will have an analytical integral.

There are some functions which are either very difficult to integrate analytically, or that simply cannot be integrated analytically. So the integral test obviously won't work if a function can't be integrated.

### Example 1

Show that the harmonic series diverges.

The harmonic series is

$$S = \sum_{n = 1}^{\infty} \, \frac{1}{n}$$

To prove convergence or divergence, we use the integral test. We integrate the function f(x) = 1/x from 1 to ∞. Here is the graph of the series and the function:

The integral is

$$\lim_{R\to\infty} \int_1^R \, \frac{1}{x} \, dx = \lim_{R\to\infty} ln(x) \bigg|_1^R$$

The integral is ln|x| evaluated as R → ∞, where ln(1) is zero:

$$= \lim_{R\to\infty} ln(R) - ln(1) \rightarrow \infty$$

So because this integral has no limit, and the function f(x) is the upper bound for the series, we conclude that the series has no upper bound and it diverges.

It's worth noticing that by the divergence test, the terms of the harmonic series approach zero. By that test, the harmonic series doesn't diverge. But remember that the divergence test can only tell us whether a series diverges. A series with decreasing terms still may not converge. One way to think about it is that the sizes of the terms, an, decrease, but just not fast enough.

### Example 2

Determine whether the series   $S = \sum_{n = 1}^{\infty} \frac{1}{n^2}$   converges.

The integral test is to find out whether the integral of f(x) = 1/x2 converges to a finite number as x→ ∞. The integral is:

$$\lim_{R\to\infty} \int_1^R \frac{1}{x^2} \, dx = \lim_{R\to\infty} -\frac{1}{x} \, \bigg|_1^R$$

This time the limit resolves to zero and the integral has a finite value of one.

$$= \lim_{R\to\infty} -\frac{1}{R} + 1 = 1$$

so this series – later we'll call it a p-series – converges.

Please note that while the integral converges to a value of one, this is not the sum of the series. In fact, if you look at the graphs in the introduction and the first example, you'll see that (1) we left out part of the series in this integral (the first box in the histogram), and (2) for the rest of the series, 1 would be an upper bound only. The sum of this series as written is actually 2.

### Example 3

Determine whether the series converges

$$S = \sum_{n = 10}^{\infty} \frac{1}{\sqrt{n - 9}}$$

First we set up the integral:

$$\lim_{R\to\infty} \int_{10}^R \frac{1}{\sqrt{x - 9}} \, dx$$

We can solve this integral by simple u-substitution:

let u = x - 9, then du = dx

to get the following integral (I've dropped the limits; I'll go back to x after evaluating the u-integral and recover the original units):

$$\int u^{-1/2} \, du = 2 u^{1/2}$$

Now the full integral looks like this:

$$\lim_{R\to\infty} \int_{10}^R \frac{1}{\sqrt{x - 9}}\, dx = \lim_{R\to\infty} 2(x - 9)^{1/2} \bigg|_{10}^R$$

Finally, as we evaluate the limits of integration, we see that the first term diverges to infinity:

$$= \lim_{R\to\infty} \, 2(R - 9)^{1/2} - 2$$

Therefore this series diverges.

### Practice problems

Determine whether the following infinite series converge or diverge using the integral test.

 1 $$\sum_{n = 1}^{\infty} \, \frac{1}{n^2}$$ 2 $$\sum_{n = 1}^{\infty} \, \frac{n + 1}{n^3}$$
 3 $$\sum_{n = 2}^{\infty} \, \frac{1}{n\cdot ln(n)}$$ 4 $$\sum_{n = 1}^{\infty} \, \frac{n}{e^n}$$

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