Calculus II ep23: The Integral Test (Nov 4, 2026)
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Overview
The lecture develops tools for deciding whether infinite series converge, beginning with the geometric and harmonic series and the necessary-term test: a nonzero term limit proves divergence, while a zero limit is inconclusive. The integral test connects a positive, decreasing function's improper integral to the convergence of its series, establishing that the harmonic series diverges and the p-series converges exactly when p > 1; the lecture closes with the Riemann zeta function and the Riemann hypothesis.
Key takeaways
- A convergent series must have terms approaching zero, so a nonzero limit for a_n proves divergence; a zero limit alone proves nothing.
- For a positive, continuous, decreasing function f on [1, infinity), sum f(n) converges exactly when the improper integral of f converges.
- The harmonic series diverges because integral from 1 to infinity of 1/x dx grows like ln(t) without bound.
- The p-series sum 1/n^p converges for p > 1 and diverges for p <= 1, providing a standard benchmark for other series.
- The integral test can establish that sum 1/n^2 converges, but it does not find its value, which is pi^2/6.
- The Riemann zeta function extends the p-series idea, and the Riemann hypothesis asserts that its nontrivial zeros lie on the line with real part 1/2.
Chapters
- A geometric series with terms ar^(n-1) converges when |r| < 1, with sum a/(1-r), and diverges when |r| > 1.
- The harmonic series, the sum of 1/n, diverges.
- The sum of 1/n^2 converges to pi^2/6, illustrating that determining a series' sum can be harder than testing its convergence.
- The lecture focuses on convergence tests because most series do not have a simple formula for their exact sum.
- If a series converges, its terms a_n must approach zero; otherwise, the series diverges.
- The proof uses partial sums: a_n = s_n - s_(n-1), and both partial-sum sequences approach the same limit when the series converges.
- For the series with terms (3n^3 - 7n)/n^3, the terms approach 3, so the series diverges by the nth-term test.
- The sequence sin(n) does not approach zero, so the series with terms sin(n) diverges.
- The nth-term test is one-way: a nonzero term limit proves divergence, but a zero limit gives no conclusion.
- The harmonic series has terms 1/n that approach zero, yet the series diverges.
- When a_n approaches zero, another convergence test is needed to decide what happens to the series.
- For f(x) = 1/x, the series terms 1/n are represented by unit-width rectangles, while the improper integral represents area under the curve.
- When f is decreasing, rectangles placed at left endpoints overestimate the area under the curve.
- Rectangles placed at right endpoints represent the series starting at n = 2 and underestimate the integral.
- Together, the comparisons bound the integral between the series starting at n = 2 and the series starting at n = 1.
- For a positive, continuous, decreasing f on [1, infinity), the series sum f(n) and improper integral from 1 to infinity of f(x) have matching convergence behavior.
- If the improper integral converges, the series converges; if the integral diverges, the series diverges.
- The integral test compares convergence only, not numerical values: the series sum and integral need not be equal.
- Checking that the function is decreasing is essential before applying the test.
- Set f(x) = 1/x, which is decreasing on [1, infinity), so the integral test applies to the harmonic series.
- The improper integral is the limit of integral from 1 to t of 1/x dx, equal to the limit of ln(t).
- Because ln(t) grows without bound as t approaches infinity, the integral diverges.
- The harmonic series sum 1/n therefore diverges by the integral test.
- For sum 1/n^2, use the decreasing function f(x) = 1/x^2 on [1, infinity).
- Writing 1/x^2 as x^(-2) gives an antiderivative of -x^(-1).
- The improper integral from 1 to infinity converges to 1, so the series sum 1/n^2 converges.
- The integral test establishes convergence but does not calculate the series sum, which is pi^2/6.
- For the p-series sum 1/n^p, the integral test shows convergence when p > 1.
- The p-series diverges when p <= 1; p = 1 gives the harmonic series.
- The example sum 1/n^3 converges because its corresponding improper integral converges.
- Memorize the classification: p-series converge for p > 1 and diverge for p <= 1.
- For real p > 1, the p-series is represented by the Riemann zeta function, ΞΆ(p) = sum 1/n^p.
- The series at p = 2 equals pi^2/6, while the value at p = 3 is not known in a comparable closed form.
- The Riemann hypothesis concerns the nontrivial zeros of the analytically continued zeta function and asserts that they have real part 1/2.
- The lecture notes that the Riemann hypothesis remains unproved and is connected to the distribution of prime numbers.
Transcript excerpt
0:00 it. All right. Uh let's talk some more about series. Just to recap what we did last time and I guess well mostly last time um we talked about two types of series. There's the geometric series and this one looks like a r to the n minus one where a and r are specific numbers and this one converges. This you got to remember this converges when absolute values are of r is less than one and it converges to a over 1 - r. That's the formula to remember. Not so complicated to remember that. And then uh it diverges when the r
0:49 is greater than one in absolute value. All right. So this is the geometric series. And we also talked just at the very end about another one called the harmonic series. and that is 1 / n. And this one diverges. There's no sort of conditions here because there's no r or a or anything like that over here. So this is the geometric series and the harmonic series. And these two examples come up uh over and over again. And so you need to be able to just kind of remember and identify these when you see them. All
1:38 right. In terms of converges versus diverges, this is something that we are going to be talking about uh a lot for the rest of the semester. I guess um in the case of the uh geometric series, we have this nice formula about what it actually converges to. Generally speaking, figuring out what a series adds up to is very difficult. So I will just say uh usually we can only decide if uh a series converges or not. Um in the case of the geometric you can actually say what it adds up to when it
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