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A sum of positive decreasing terms and the integral of the function behind them share a verdict, which hands the improper integrals of unit 1 to series; and the family $\sum n^{-p}$, which converges exactly when $p > 1$.
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 can state the integral test with all three of its hypotheses, check them on a given series before quoting it, and read the series' verdict off the improper integral. You can apply the $p$-series rule directly, placing the boundary case $p = 1$ on the diverging side, and say why the integral's value is not the series' sum.
You can evaluate $\int_1^{\infty} x^{-p}\,dx$ as a limit and you know it converges exactly when $p > 1$. This lesson hands that result to series: a sum of positive decreasing terms and the area under the curve those terms come from are close enough that they stand or fall together.
The integral test: if $f$ is positive, decreasing and continuous on $[1, \infty)$ and $a_n = f(n)$, then $\sum a_n$ and $\int_1^{\infty} f$ either both converge or both diverge.
$p$-series: $\sum \dfrac{1}{n^{p}}$.
Harmonic series: the $p$-series with $p = 1$, the boundary case, and it diverges.
Tail: everything past some fixed term. Convergence is decided entirely by the tail.
Draw the terms of $\sum a_n$ as rectangles of width $1$ and height $a_n$, and draw the curve $y = f(x)$ through their corners. Because $f$ is decreasing, the rectangles can be lined up to sit entirely under the curve or entirely over it, so the total area of the blocks and the area under the curve differ by at most one block. One of them is finite exactly when the other is.
The integral test. If $f$ is positive, decreasing and continuous on $[1, \infty)$ and $a_n = f(n)$, then
$$\sum_{n=1}^{\infty} a_n \text{ converges} \iff \int_1^{\infty} f(x)\,dx \text{ converges.}$$
Three hypotheses, and all three are load-bearing. Positive, so the blocks do not cancel. Decreasing, so they can be trapped against the curve. Continuous, so the integral exists at all. Every convergence test is a theorem with conditions attached, and a test quoted where its conditions fail has not been applied — it has been guessed with. Naming the condition you checked is part of the answer, and on an examination it is part of the marks.
The $p$-series rule falls straight out. Apply the test to $f(x) = x^{-p}$:
$$\sum \frac{1}{n^{p}} \text{ converges} \iff p > 1.$$
The boundary is at $p = 1$ because that is the single exponent whose antiderivative is $\ln x$ rather than a power of $x$, and $\ln b$ grows without limit. So the harmonic series diverges, $\sum n^{-2}$ converges, and $\sum n^{-1/2}$ diverges.
A verdict, never a sum. $\int_1^{\infty} x^{-2}dx = 1$, and $\sum n^{-2} = \tfrac{\pi^2}{6} \approx 1.645$. The two numbers are not equal and were never going to be; only the answer to does it settle crosses between them.
Another way: picture
Picture a staircase of blocks descending to the right, and a smooth slide drawn through their top-left corners. Slide the blocks one step right and they all drop below the curve; leave them where they are and they all poke above it. Either way the blocks and the region under the curve differ by no more than the first block, so a finite area means a finite sum and an infinite one means an infinite sum.
Another way: steps
| $p$ | The series | The integral |
|---|---|---|
| $\tfrac12$ | diverges | diverges |
| $1$ | diverges | diverges |
| $\tfrac32$ | converges | converges |
| $2$ | converges | converges |
| $3$ | converges | converges |
The two columns agree in every row, and they have to. The only line worth memorising separately is the second one: $p = 1$ is the harmonic series and it falls on the diverging side, which is not obvious from anything about the terms themselves.
Reading the integral's value as the sum. They are different numbers. The test transfers a verdict and nothing more.
Skipping the hypotheses because the function 'looks fine'. $\dfrac{\sin x}{x}$ looks fine and is not positive; $\dfrac{\ln x}{x}$ looks fine and is not decreasing at first. Both are ordinary-looking functions that need a sentence of justification before the test is quoted.
Putting $p = 1$ on the wrong side. The rule is $p > 1$, strictly. The harmonic series is the case people remember wrongly, and it is also the one that comes up most.
$\displaystyle\sum \frac{1}{n^{3}}$: the function $x^{-3}$ is positive, decreasing and continuous on $[1, \infty)$.
Hypotheses first.
$\displaystyle\int_1^{\infty} x^{-3}dx = \lim_{b\to\infty}\left[\frac{x^{-2}}{-2}\right]_1^b = \frac12$, which is finite.
So the integral converges.
So the series converges. Quoting the $p$-series rule with $p = 3 > 1$ would have reached the same verdict in one line — and the sum is not $\tfrac12$.
$\displaystyle\sum \frac1n$: $f(x) = \tfrac1x$ is positive, decreasing and continuous on $[1, \infty)$.
The test applies.
$\displaystyle\int_1^{b} \frac{dx}{x} = \ln b$, which grows without limit.
A logarithm, not a power.
So the series diverges. It does so extraordinarily slowly — a hundred terms reach about $5.19$ — which is why it is so easy to believe it settles.
$f(x) = \dfrac{1}{x(\ln x)^2}$ is positive, decreasing and continuous on $[2, \infty)$, so the test applies from there.
Starting at 2 is allowed; only the tail matters.
Substituting $u = \ln x$ turns the integral into $\displaystyle\int_{\ln 2}^{\infty} \frac{du}{u^{2}}$.
A $p$-integral with $p = 2$.
That converges, so the series converges — though only just: the same series with $(\ln n)^1$ instead diverges.
For each exponent $p$, say what $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{p}}$ does and what $\displaystyle\int_1^{\infty} \dfrac{1}{x^{p}}\,dx$ does.
| The series | The integral | |
|---|---|---|
| $p = 3$ | ||
| $p = 1$ | ||
| $p = \dfrac{1}{3}$ |
Put the steps of applying the integral test to $\displaystyle\sum \dfrac{1}{n^{2}}$ into the order you do them.
Number the steps in order (write the number in the box):
Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{\frac{1}{2}}}$ converge or diverge?
Evaluate $\displaystyle\int_1^{\infty} \dfrac{1}{x^{3}}\,dx$.
Answer:
Match each series to what the integral test's hypotheses do for it.
| The terms are not all positive | The terms increase rather than decrease | All three hypotheses hold from the start | All three hold only from some point onwards | |
|---|---|---|---|---|
| $\sum \dfrac{(-1)^n}{n}$ | ||||
| $\sum n^2$ | ||||
| $\sum \dfrac{1}{n^3}$ | ||||
| $\sum \dfrac{\ln n}{n}$ |
Select every series the integral test may be quoted for, allowing it to be applied from any starting point.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For each exponent $p$, say what $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{p}}$ does and what $\displaystyle\int_1^{\infty} \dfrac{1}{x^{p}}\,dx$ does.
| The series | The integral | |
|---|---|---|
| $p = 2$ | ||
| $p = 1$ | ||
| $p = \dfrac{1}{2}$ |
You can use the integral test and the $p$-series rule. Without looking: what are the three hypotheses, for which $p$ does $\sum n^{-p}$ converge, and does the integral's value give you the sum?
9. Your turn: $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}$, step 3