Integral Test, Approximation

In a typical series, its terms a_k are given by some formula. Usually this formula makes sense not only for integers k from the indexing range, but also for real numbers between them; that is, the formula defines a function f on the interval [n₀,∞) and the individual terms of the series are just some special values of this function, a_k = f (k).

Is there any way to relate the behavior of such a series to the behavior of this function f? Under some circumstances it is possible.

Theorem (Integral test, IT).
Assume that f is a positive, non-increasing function on an interval [n₀,∞) for some integer n₀. For all integers k ≥ n₀ define a_k = f (k).
Then the series  a_k  converges if and only if the improper integral    converges.
Moreover, then we have

Why should something like this be true? The key observation is this: When we see a term a_k of this series, we may interpret it as the area of a rectangle whose height is a_k and the base is 1. This we can do with any number, of course, the real trick is finding a suitable rectangle to apply this idea to. We decide to use the above picture and place this particular rectangle is such a way that its upper left corner coincides with the dot marking a_k. Then its base lies on the x-axis. We do this with all terms of the series.

Now look again at the rectangle associated with a_k. Since f is non-increasing and passes through the upper-left corner of this rectangle, it follows that the integral of f between k and k + 1, that is, the area under f between these two numbers, cannot exceed the area of our rectangle, which is a_k. In this way we establish a relation between a certain definite integral of f and the term a_k. We can do it for every term of the series and then sum these inequalities up, then we obtain

Thus we see that if the integral is divergent, that is, infinite, then the series must also give infinity. Conversely, if the series converges, then the integral must also converge (here we heavily use the fact that f > 0, so the integral cannot run away from us to negative parts). This establishes one direction, one estimate for the series. In fact, a proper mathematical proof looks almost exactly the same, it just needs to be polished up a bit, especially one should worry about convergence where appropriate (pass to partial sums etc.). However, the main idea is as we explained it.

It remains to establish the other direction, that is, an upper estimate for the series. This time we position our rectangles so that the points of the series are at their upper right corners. In this way the rectangle associated with a_n0 would go outside of f, so we will disregard it.

The area a_k is now bounded from above by the integral of f from k − 1 to k.

Now we just add a_n0 to both sides of this inequality and we get the desired estimate. We also see that if the series gives infinity, then the integral must be also infinite. On the other hand, if the integral converges, then taking into account that the series cannot run away from us to negative numbers we conclude that it must also converge.

Note that the last inequality in the calculation shows that we also have the following upper estimate for our sum:

Indeed, many authors (perhaps even most) actually put this upper estimate in their version of Integral test theorem, since it comes directly out of the proof. However, it never gives a better estimate than our version and in most cases it actually works worse, so we will not loose anything by ignoring it here.

Remark: In the theorem we used n₀ as the starting point for our indexing. However, if we are not interested in any estimate, just in the fact of convergence or divergence, then we know that this starting index becomes irrelevant. The integral test can be then simplified like this: A series  ∑ a_k  converges if and only if the integral of the corresponding f from some suitable K to infinity also converges (assuming that f is positive and non-increasing, otherwise the test cannot be used).

Example: We will prove now that the series converges.

The formula in the series also defines a function f (x) = 1/x² that is positive and decreasing for instance on [1,∞). Therefore we can use the Integral test. Since we know that this function has a convergent integral on this interval, by IT it follows that also the given series converges.

Remark: We should actually remember the convergence of this series, we usually refer to the p-test, see Theory - Introduction - Important examples. This p-test itself is proved by a straightforward application of the Integral test. We simply replace 2 by p in the above argument and then look at the convergence of the integral (at infinity) of the function f (x) = 1/x^p, see also Properties of improper integrals in Integrals - Theory - Improper integrals.

A nice example of using the Integral test can be found in Methods Survey and others in Solved Problems - Testing convergence, namely this problem, see also this problem.

The two inequalities in the Integral test above allow us to also approximate the sum in case we are unable to find its precise value (which is the usual situation). Note that the difference between the lower and the upper estimate is exactly a_n0. We know that the terms of a convergent series must go to zero, which tells us that we get a more precise answer if we apply this estimate only to the tail of the given series. Look at the following example.

We proved above that the series converges. Now we will try to estimate its sum when indexing starts from 1. Thanks to some advanced theory we actually know what it is, see the appropriate example in section Important examples in Theory - Introduction to series, but that result was obtained by a very specific trick. Here we have a relatively low-tech method with much greater potential for application.

According to the Integral test we have

We have an estimate, but it is not too precise. Given that the result is about 1, the error of 1 is way too large. Let's say that we would want to have an error at most 0.005. Since the error is given by 1/n₀², we see that in order to make it small enough we have to sum up starting from 15 (or further). What about the beginning of the series that we skip in this way? That beginning is actually a finite sum, so we can evaluate it precisely and add to all parts of the two inequalities.

We might guess that the sum is about 1.645 and we will not be far off. The method that we used can be expressed in general by this formula:

The difference between the upper and lower bound is a_K which for a convergent series goes to 0, so we get estimates of arbitrary precision just by taking sufficiently large integer K.

Comparison tests
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