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The Integral Test and p-Series

Calculus 2 · Chapter 7: Convergence Tests for SeriesSection 7.1 · lesson 32 of 549 min · free

← Properties of Convergent SeriesNext lesson: The Comparison and Limit Comparison Tests →

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In this lesson

Trading a sum for an integral you already know how to do.

  • Check the three hypotheses before applying the test
  • Evaluate the matching improper integral to get the verdict
  • Memorise the p-series rule: converges exactly when p>1
  • Reach series the p-test alone cannot classify

Transcript

Everything said in the video, section by section (1,298 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 7.1

The Integral Test and p-Series

Chapter Six gave you exact sums for two families. Geometric, and telescoping. Everything else needs a different approach. A test. A way to settle whether a series converges, without ever computing its sum. And the first such test exploits a picture that's been available since the very first integral you drew. A series of decreasing positive terms is a stack of unit-width rectangles. And an integral is the area under the curve those rectangles approximate. So if you can compute the integral, you can trap the series.

The Idea

A Sum Sandwiched Against an Integral

Look at the two pictures, because the whole test lives in them. On the left, rectangles drawn under the curve. Each rectangle has width one, and height equal to the function value at the right endpoint of its interval. Because the function is decreasing, every rectangle sits below the curve. So the total area of those rectangles is at most the integral. Which means: if the integral is finite, the series is too. On the right, rectangles drawn over the curve. Same width, but now the height is the value at the left endpoint. Every rectangle pokes above the curve. So their total area is at least the integral. Which means: if the integral is infinite, the series is as well. Put those two together, and the series is sandwiched against the improper integral. Whatever the integral does, the series does the same.

The Test

Three Hypotheses, Then Integrate

Here's the formal statement. Suppose a sub n equals f of n, where on the interval from one to infinity, the function f is positive, continuous, and decreasing. At least eventually — a finite number of early terms never matters, as section six point five established. Then the series and the improper integral converge or diverge together. Applying the test is a clean two-stage process. First, confirm the three hypotheses for f on the relevant interval. Positive. Continuous. Decreasing. Skip that check and the test may give you a wrong answer. Second, evaluate the improper integral, using the limit definition from Chapter Two. Replace the infinite upper limit with t. Integrate. Take the limit as t goes to infinity. Whatever the integral does, the series does the same. One warning, and it matters. The test transfers only the verdict. Not the value. A convergent series almost never equals its integral. So don't report the integral as the sum.

Worked Example 1

A Convergent Series

Let's do one. Determine whether the sum of one over n squared plus one converges. Let f of x be one over x squared plus one. Check the hypotheses. On the interval from one to infinity, it's positive — the denominator is always positive. It's continuous — no zeros in the denominator. And it's decreasing, since the denominator grows. So the Integral Test applies. Evaluate the improper integral. The antiderivative of one over x squared plus one is arctangent x. Replace infinity with t. You get the limit of arctangent t, minus arctangent one. Arctangent t goes to pi over two. And arctangent one is pi over four. So the integral converges, to pi over four. By the Integral Test, the series converges as well. Now notice the subtlety I flagged earlier. The series converges. But its sum is not pi over four. The test transfers the yes-or-no verdict. Not the numerical value. There is a bonus, though, unique to this test. Because the leftover tail of the series is sandwiched between two integrals, the remainder after N terms is at most the integral from N to infinity. That remainder estimate lets you approximate a convergent sum to a guaranteed accuracy — by adding enough terms. The same picture that decides convergence also controls the error of a partial-sum approximation.

The Headline Result

The p-Series Test

The Integral Test's most important payoff is a complete, memorizable verdict on one specific family. The p-series. The sum of one over n to the p. These serve as the universal yardstick against which other series are later compared. So their behavior must be known cold. Here's the result. The p-series converges if p is greater than one. And diverges if p is at most one. And notice — that mirrors exactly the p-integral result from Chapter Two. The integral from one to infinity of x to the minus p converges if and only if p exceeds one. Which is precisely why the Integral Test delivers this. One over n to the p matches the function one over x to the p. Whose improper integral has a known, simple convergence threshold. The Integral Test passes that threshold straight to the series. Now — the boundary case. p equals one is the harmonic series. Which diverges. The classic warning that terms shrinking to zero do not guarantee convergence. Just above the threshold, p equals one point zero one converges. Just below, p equals zero point nine nine diverges. The dividing line at p equals one is razor-sharp. And the most common error in this entire chapter is treating p equals one as convergent.

Worked Examples 2 and 3

Reading Off, and Reaching Further

Let's classify some on sight. The sum of one over n cubed. Here p is three, greater than one. So it converges. The sum of one over root n. Since root n is n to the one half, this is a p-series with p equal to one half. One half is at most one. So it diverges. And the sum of one over n. That's the harmonic series. p equals one. Diverges. Recognizing a p-series on sight — and recalling that the cutoff is strictly p greater than one — resolves these instantly. No integral to evaluate at all. Now one where the p-series test doesn't directly apply. Determine whether the sum of one over n log n converges, starting at n equals two. This is not a p-series. But f of x equals one over x log x is positive, continuous, and decreasing on the interval from two to infinity. So the Integral Test applies. Using the substitution u equals log x — from Chapter One — the antiderivative is log of absolute log x. Evaluate. You get the limit of log log t, minus log log two. And log log t goes to infinity. Slowly. But it goes. So the integral diverges, and the series diverges. This example shows the Integral Test reaching cases the p-series test alone cannot classify. The terms here shrink even more slowly than any p-series with p greater than one. Yet still not fast enough to converge.

Key Takeaways

What to Carry Forward

Here's something to think about, and it goes to the heart of the p-series threshold. The harmonic series, with p equal to one, diverges. But the sum of one over n to the one point zero zero zero one converges. Both have terms that shrink to zero at almost identical rates. You could not tell them apart from a table of values. So using the integral picture, explain what is fundamentally different about the area under one over x, versus one over x to the one point zero zero zero one, out to infinity. And why does an infinitesimal change in the exponent flip the verdict entirely? Five takeaways. The Integral Test applies when f is positive, continuous, and decreasing — eventually — with a sub n equal to f of n. Then the series and the integral converge or diverge together. The test transfers only the verdict, not the value. A convergent series almost never equals its integral. Evaluate the improper integral by the limit method of Chapter Two. The test does the rest. The p-series converges if and only if p is greater than one. That's the single most-used benchmark in this chapter. And the harmonic series, with p equal to one, diverges. Shrinking terms alone never guarantee convergence. Next up is section seven point two — the Comparison and Limit Comparison Tests. Which measure an unknown series against exactly these benchmarks.

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FAQ

Questions students ask us

About The Integral Test and p-Series

What is the integral test for series?
The Integral Test says that if a_n = f(n), where f is positive, continuous, and decreasing on [1,\infty) (at least eventually), then \sum a_n and \int_1^\infty f(x)\,dx either both converge or both diverge. It works because unit-width rectangles drawn under the curve keep the series below the integral, while rectangles drawn over the curve keep it above. Always verify all three hypotheses before integrating.
When does a p-series converge?
A p-series \sum 1/n^p converges exactly when p>1 and diverges when p\le 1. This follows from the Integral Test, since \int_1^\infty x^{-p}\,dx converges only for p>1. For example, \sum 1/n^3 converges, while \sum 1/\sqrt{n} has p=1/2 and diverges. The most common mistake is treating p=1 as convergent. That case is the harmonic series \sum 1/n, which diverges even though its terms go to zero.
Does the integral test give the sum of a convergent series?
No, the Integral Test tells you only whether the series converges, not what it sums to. For \sum 1/(n^2+1), the integral \int_1^\infty \frac{dx}{x^2+1} = \lim_{t\to\infty}(\arctan t - \arctan 1) = \pi/2 - \pi/4 = \pi/4, so the series converges, but its sum is not \pi/4. The integral does give a useful error bound: the remainder after N terms is at most \int_N^\infty f(x)\,dx.
Does the series of 1/(n ln n) converge?
No, \sum_{n=2}^\infty \frac{1}{n\ln n} diverges, and the Integral Test shows this even though it is not a p-series. The function f(x)=\frac{1}{x\ln x} is positive, continuous, and decreasing on [2,\infty). Substituting u=\ln x gives the antiderivative \ln|\ln x|, so the integral equals \lim_{t\to\infty}(\ln\ln t - \ln\ln 2), which is infinite. The terms shrink more slowly than those of any p-series with p>1.

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