5.3 The Divergence and Integral Tests

Big idea

Only a handful of series can be summed exactly. For everything else the realistic goal is a verdict: converges or diverges, with no value attached. This section supplies the first two verdict-producing tools, one cheap and one powerful.

The cheap one is the divergence test. Look at the terms alone. If they do not approach zero, the running totals keep receiving contributions of roughly the same size forever and cannot converge. It costs one limit, it should be the first thing you try on any unfamiliar series, and it settles a surprising number of cases immediately. Its weakness is equally definite: when the terms do go to zero it tells you nothing at all.

The powerful one is the integral test, and it rests on a picture. A series with positive decreasing terms can be drawn as a stack of rectangles of width $1$ and heights $a_1, a_2, a_3, \dots$. If the terms come from a decreasing function $f$ with $a_n = f(n)$, those rectangles sit just above or just below the graph, depending on which corner you anchor them to. The total rectangle area is the series, the area under the curve is an improper integral, and each traps the other within one term. So the two are finite together and infinite together.

That comparison converts a question about an infinite sum into a question about an improper integral, which is a technique you already own from the integration chapters. It immediately settles the p-series family, which then becomes the benchmark that the comparison tests in the next section measure everything else against. It also does something the other tests cannot: because the picture bounds the tail, it estimates how far a partial sum is from the true value.

Decoder

If the terms of a series are the values at the integers of a positive continuous decreasing function, then the series and the improper integral of that function either both converge or both diverge.

Every adjective in that sentence is a hypothesis, and the hypotheses are the reason the picture works. Positive keeps the partial sums increasing, so they either climb to a finite ceiling or to infinity with nothing in between. Decreasing is what makes each rectangle fit cleanly against the curve. Continuity is what makes the integral exist in the first place.

The conclusion is about agreement, not equality. The series and the integral almost never have the same value, and you should never report the integral as the sum. They share a verdict and nothing more. What the integral does give quantitatively is a bound on the leftover after a partial sum, and that is a separate statement proved from the same picture.

Two practical notes. The hypotheses only need to hold from some index onward, since changing finitely many terms cannot change convergence, so a function that decreases only after $x = 4$ is fine if you start the integral at $4$. And if the terms are not eventually positive or not eventually decreasing, the test does not apply and its conclusion is not available to you, even if the integral happens to converge.

Definitions and results

The divergence test. If $\lim_{n \to \infty} a_n \neq 0$, or the limit does not exist, then $\sum a_n$ diverges. This is the only conclusion the test can produce.

The limitation, stated plainly. If $\lim_{n \to \infty} a_n = 0$, the test yields nothing and another tool is required. Both $\sum \frac1n$ and $\sum \frac{1}{n^2}$ pass this screen, and only one of them converges.

The integral test. Suppose $f$ is continuous, positive, and decreasing on $[N, \infty)$ and $a_n = f(n)$ for $n \ge N$. Then

$$ \sum_{n=N}^{\infty} a_n \quad \text{and} \quad \int_{N}^{\infty} f(x)\,dx $$

converge together and diverge together.

Why the picture works. Rectangles of width $1$ drawn with their left corners on the curve lie above it, so $\int_{N}^{M+1} f \le \sum_{n=N}^{M} a_n$. Drawn with their right corners on the curve they lie below it, so $\sum_{n=N+1}^{M} a_n \le \int_{N}^{M} f$. Each side is finite exactly when the other is.

The p-series. For $p > 0$,

$$ \sum_{n=1}^{\infty}\frac{1}{n^{p}} \quad \text{converges if } p > 1 \text{ and diverges if } p \le 1 $$

The case $p = 1$ is the harmonic series and it diverges, which is the boundary you should memorize by its integral: $\int_1^{\infty} x^{-1}dx$ grows like $\ln x$, without bound but barely.

Remainder estimate. If the integral test applies and $S = \sum_{n=1}^{\infty} a_n$ with partial sum $S_M$, then the remainder $R_M = S - S_M$ satisfies

$$ \int_{M+1}^{\infty} f(x)\,dx \le R_M \le \int_{M}^{\infty} f(x)\,dx $$

This brackets the error, which no other test in this chapter does for a general positive series.

What the integral test cannot do. It never produces the sum. It does not apply to series whose terms change sign or fail to decrease. And it is useless in practice whenever the antiderivative is out of reach, which happens often: the test is valid for $\sum \frac{1}{n^3 + 1}$ but evaluating $\int_1^{\infty}\frac{dx}{x^3+1}$ is an unpleasant partial-fractions exercise, so a comparison is the better move.

Worked examples

One series the divergence test settles and one it does not

Consider $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2n + 7}$ and $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n \ln n}$.

For the first, divide numerator and denominator by $n$: the terms tend to $\frac12$, which is not $0$. The series diverges, and one limit was the entire argument.

For the second, the terms tend to $0$, since both factors in the denominator grow. The divergence test is therefore inconclusive here, and reporting “the terms go to zero so it converges” would be wrong. The next example shows what actually happens.

The integral test on a borderline case

Decide $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$.

Take $f(x) = \frac{1}{x \ln x}$ on $[2, \infty)$. It is continuous there, positive there since $\ln x > 0$ for $x \ge 2$, and decreasing there because both $x$ and $\ln x$ increase, making the denominator increase. All three hypotheses hold, so the test applies.

Substitute $u = \ln x$, $du = \frac{dx}{x}$:

$$ \int_{2}^{b}\frac{dx}{x \ln x} = \int_{\ln 2}^{\ln b}\frac{du}{u} = \ln(\ln b) - \ln(\ln 2) $$

As $b \to \infty$, $\ln(\ln b) \to \infty$, so the integral diverges and therefore so does the series. The terms of this series are smaller than $\frac1n$ for every $n \ge 3$, and it still diverges. That is how narrow the boundary between convergence and divergence is.

The p-series family from one integral

For $p \neq 1$ and $b > 1$,

$$ \int_{1}^{b} x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_{1}^{b} = \frac{b^{1-p} - 1}{1 - p} $$

If $p > 1$ then $1 - p < 0$, so $b^{1-p} \to 0$ and the integral converges to $\frac{1}{p-1}$. If $p < 1$ then $1 - p > 0$ and $b^{1-p} \to \infty$, so it diverges. The case $p = 1$ gives $\ln b \to \infty$, also divergence. Since $x^{-p}$ is continuous, positive, and decreasing on $[1,\infty)$ for every $p > 0$, the integral test transfers each verdict to the series.

So $\sum \frac{1}{n^{3/2}}$ converges, $\sum \frac{1}{\sqrt n}$ diverges, and $\sum \frac{1}{n^{1.001}}$ converges even though it looks indistinguishable from the harmonic series for any number of terms you could ever add.

Bracketing the error in a partial sum

Estimate the remainder of $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ after ten terms.

Here $f(x) = x^{-2}$ satisfies every hypothesis on $[1, \infty)$. The two integrals are

$$ \int_{11}^{\infty}\frac{dx}{x^2} = \frac{1}{11}, \qquad \int_{10}^{\infty}\frac{dx}{x^2} = \frac{1}{10} $$

so $0.0909 \le R_{10} \le 0.1$. Adding the first ten terms gives $S_{10} \approx 1.5498$, so the true sum lies between about $1.6407$ and $1.6498$.

Check this against the known value $\frac{\pi^2}{6} \approx 1.6449$, which sits inside the bracket, and the actual remainder is about $0.0952$, comfortably between $\frac{1}{11}$ and $\frac{1}{10}$. Notice how slow the convergence is: ten terms and the answer is only good to about one decimal place. Knowing a series converges is not at all the same as being able to compute it.

Practice

First the screen. Take the limit of the terms, declare divergence when it is nonzero, and state that the test is inconclusive when the limit is zero rather than guessing.

Then the integral test and its most important consequence. Verify the three hypotheses before integrating, use the p-series result as a shortcut once it is established, and bracket a remainder when asked for accuracy.

Quiz

Five items on the divergence test and its limits, the hypotheses and use of the integral test, p-series, and remainder estimates.