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Ruling a series out by its terms, and tying a series to an improper integral by rectangles of width one.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to apply the divergence test in the one direction it runs and refuse to read it backwards, state the integral test with its three hypotheses and say what each of them does in the proof, derive the $p$-series test from it and place the borderline at $p = 1$, recognise the series the integral test cannot be applied to at all, distinguish an inconclusive test from a verdict of divergence, and use the same rectangles to bound the remainder of a truncated series.
Geometric and telescoping series can be summed because their partial sums have closed forms. Almost nothing else does. $\sum \dfrac{1}{n^2}$ has the sum $\dfrac{\pi^2}{6}$, which took a century to find and is not obtainable by any method in this course; $\sum \dfrac{1}{n^3}$ has no known closed form at all.
So the question changes. Instead of asking what a series sums to, we ask whether it sums to anything — and that turns out to be answerable, often in one line, by tests that never produce the number.
A test is a theorem that concludes convergence or divergence from a checkable property of the terms. Every test has hypotheses, and a test whose hypotheses fail has not been applied — it has been guessed with.
A test is inconclusive on a series when its hypotheses hold and its conclusion does not fire. That is not a verdict of divergence; it is silence, and another test is needed.
The remainder $R_N = \sum_{k > N}a_k$ is what is discarded when a series is truncated. A test that bounds the remainder gives an approximation with an error bar rather than a number with a hope attached.
The divergence test. If $a_n \not\to 0$ then $\sum a_n$ diverges.
It is the first thing to check and it costs nothing. It runs in one direction only: terms tending to zero establishes nothing whatever. Terms going to zero is necessary and nowhere near sufficient. The harmonic series has terms going to zero and diverges; that single example is why the divergence test is stated in one direction only and can never certify convergence.
The integral test. Let $f$ be positive, decreasing and continuous on $[1,\infty)$, with $a_n = f(n)$. Then $$\sum_{n=1}^{\infty}a_n \text{ converges} \iff \int_1^{\infty}f(x)\,dx \text{ converges}.$$
The three hypotheses all do work. Positive makes the partial sums increase; decreasing makes the rectangles comparable with the curve; continuous makes the integral exist. Drop decreasing and the comparison fails in both directions.
It does not give the sum. $\sum \frac{1}{n^2}$ and $\int_1^\infty x^{-2}dx$ both converge, to $\frac{\pi^2}{6} \approx 1.645$ and to $1$ respectively. The test says they agree on the verdict, never on the value.
The $p$-series test, which is its most-used consequence: $$\sum_{n=1}^{\infty}\frac{1}{n^p} \text{ converges} \iff p > 1.$$ Same borderline as the $p$-integral, for exactly this reason.
The remainder estimate. If the test applies, $$\int_{N+1}^{\infty}f \le R_N \le \int_{N}^{\infty}f.$$ The same rectangles, read as a bound rather than as a verdict.
Another way: picture
The curve $y = f(x)$ with unit-width rectangles drawn two ways: heights $f(n)$ placed to the right of each integer, so they sit under the curve, and heights $f(n)$ placed to the left, so they sit over it. The sum is the total rectangle area and the integral is the area under the curve, and the two pictures trap the sum between $\int_1^\infty f$ and $f(1) + \int_1^\infty f$. Every statement of this lesson is one of those two pictures.
Another way: steps
The integral test settles $\sum \frac1n$ by $\int_1^\infty \frac{dx}{x} = \lim \ln b = \infty$: divergent. It settles $\sum \frac{1}{n^{1.01}}$ by $\int_1^\infty x^{-1.01}dx = 100$: convergent.
Between those two exponents nothing visible changes about the terms. At $n = 10^6$ the two terms are $10^{-6}$ and $0.87 \times 10^{-6}$. No computation distinguishes them, and the partial sums of the harmonic series pass $14$ only at $n \approx 10^6$ and climb like $\ln N$ forever after.
The test goes further. $\sum \dfrac{1}{n\ln n}$ has terms much smaller than $\frac1n$, and it still diverges: $\int \frac{dx}{x\ln x} = \ln\ln x \to \infty$. But $\sum \dfrac{1}{n(\ln n)^2}$ converges, since $\int \frac{dx}{x(\ln x)^2} = -\frac{1}{\ln x}$ settles. So there is no slowest divergent series and no fastest convergent one: between any two you can insert another.
That is why the subject needs theorems. There is no threshold of smallness that decides convergence, and no amount of numerical evidence that substitutes for a proof.
Reading the divergence test as a convergence test. The single most expensive error in the unit. $a_n \to 0$ proves nothing.
*Applying the integral test without checking decreasing.* The hypothesis is what makes the rectangles comparable. A term like $\frac{2 + \sin n}{n^2}$ is positive and not monotone, and the test does not apply to it.
Reporting the integral's value as the series' sum. They agree on the verdict and almost never on the value.
Trying to integrate a factorial. $n!$ is a function on the whole numbers. There is nothing to integrate.
Insisting the hypotheses hold from $n = 1$. Convergence depends on the tail, so "decreasing for $n \ge 7$" is enough; apply the test from there.
Treating an inconclusive test as a verdict of divergence. Silence is not a no.
Forgetting that the borderline diverges. $p = 1$ fails, for the series as for the integral.
Every test in this unit is an implication, and implications are one-directional. The divergence test says terms not to zero $\implies$ divergent. Learners routinely use it as though it also said terms to zero $\implies$ convergent, which would make the harmonic series convergent and the whole subject unnecessary.
The deeper habit behind the error is treating a test as a machine that returns a verdict for any input. It is not. It is a theorem with hypotheses and a conclusion, and there are three outcomes: the conclusion fires, the conclusion does not fire and the test is silent, or the hypotheses fail and the test was never applicable. Only the first is a result.
So every answer in this unit should have three parts: which test, which hypotheses were checked, and what the conclusion was. "It converges by the integral test" is not an answer if $f$ was never shown to be decreasing — and that particular omission is not academic, because non-monotone terms are exactly where the test breaks.
The habit pays for itself in the next lessons, where the ratio test is genuinely inconclusive on an entire family of series ($p$-series, every one of them) and a learner who reads inconclusive as divergent will get all of them wrong.
$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$: the terms are smaller than $\frac1n$, so no comparison with the harmonic series can prove divergence.
Comparison points the wrong way here.
$f(x) = \dfrac{1}{x\ln x}$ is positive and decreasing on $[2,\infty)$, so the test applies. Substituting $u = \ln x$: $\displaystyle\int_2^{b}\frac{dx}{x\ln x} = \ln\ln b - \ln\ln 2 \to \infty$.
Hypotheses checked, then the integral.
So the series diverges. Change the exponent: $\sum \dfrac{1}{n(\ln n)^2}$ gives $\int \frac{du}{u^2}$, which converges. One logarithm apart, opposite verdicts, and nothing in the terms shows it.
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3}}$: $f(x) = x^{-3}$ is positive and decreasing, and $\int_1^\infty x^{-3}dx = \tfrac12$ converges, so the series does.
The test gives the verdict.
Add ten terms: $s_{10} \approx 1.19753$. What is missing? $R_{10} \le \displaystyle\int_{10}^{\infty}x^{-3}dx = \dfrac{1}{200} = 0.005$.
The same rectangles give the bound.
So the sum is between $1.1975$ and $1.2026$. The true value is $1.20206\ldots$ — Apéry's constant, which has no known closed form. The series cannot be summed and can be located, which is usually what is actually wanted.
Terms first: $\dfrac{n}{n^2+1} \to 0$, so the divergence test is silent and says nothing either way.
Always check, and never over-read the answer.
$f(x) = \dfrac{x}{x^2+1}$ is positive on $[1,\infty)$ and decreasing there (its derivative is $\dfrac{1 - x^2}{(x^2+1)^2} \le 0$ for $x \ge 1$), so the test applies.
Check decreasing rather than assuming it.
$\displaystyle\int_1^{b}\frac{x\,dx}{x^2+1} = \tfrac12\ln(b^2+1) - \tfrac12\ln 2 \to \infty$, so the series diverges — although its terms tend to zero.
Which test settles $\sum \left(\dfrac{n}{2n + 1}\right)^{n}$ most directly?
For which $p$ does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ converge? Give the set of $p$ as an interval.
This task has no paper form; do it on a device.
Build the proof that if $f$ is positive and decreasing on $[1,\infty)$ and $\displaystyle\int_1^{\infty}f$ converges, then $\displaystyle\sum_{n=1}^{\infty}f(n)$ converges.
This task has no paper form; do it on a device.
For each series, say whether the integral test may be applied.
| Yes — positive, decreasing and integrable | No — the terms are not positive | No — there is no function of a real variable to integrate | |
|---|---|---|---|
| $\sum \dfrac{1}{n^{2} + 7}$ | |||
| $\sum \dfrac{(-1)^{n}}{n}$ | |||
| $\sum \dfrac{7^{\,n}}{n!}$ | |||
| $\sum \dfrac{1}{n\ln n}$ |
For each series, give the exponent $p$ and the verdict.
| Exponent $p$ | Verdict | |
|---|---|---|
| $\sum \dfrac{1}{n^{4}}$ | ||
| $\sum \dfrac{1}{n}$ | ||
| $\sum \dfrac{1}{\sqrt{n}}$ |
For $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{2}}$, the first $9$ terms are added and the rest discarded. Bound what was discarded, using $\displaystyle\int_{9}^{\infty}x^{-2}dx$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Which test settles $\sum \dfrac{(-1)^n}{n}$ most directly?
You can settle a positive decreasing series with an integral and report which hypotheses you checked. Say in your own words why the harmonic series makes the converse of the divergence test impossible. Next: settling a series by comparing it with one you already know.
10. Your turn: does $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^{2} + 1}$ converge?, step 3