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Convergence settled by inequality rather than evaluation: below a convergent yardstick, above a divergent one, or by the limit of the ratio.
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 choose a $p$-integral yardstick by keeping the dominant terms at the end in question, state the comparison test with its non-negativity hypothesis and prove it from monotone convergence, recognise which two of the four possible pairings settle nothing and why, apply limit comparison when no inequality is obvious and say what the ratio's limit is and is not, use a comparison to bound the value of a tail as well as to decide its convergence, and give one comparison per trouble spot.
The last two lessons evaluated improper integrals by finding an antiderivative and taking a limit. That works for powers, for exponentials, and for very little else.
$\displaystyle\int_1^\infty e^{-x^2}dx$ has no elementary antiderivative. Neither does $\int_1^\infty\frac{dx}{\sqrt{1 + x^3}}$. Both converge, and the question of whether they converge is answerable without evaluating anything — which is the whole point of this lesson, and the pattern that the entire series unit will follow.
A yardstick is an integral whose behaviour you already know: here, a $p$-integral or an exponential. Comparison settles a new integral by trapping it against one.
The dominant term of a polynomial expression at infinity is its highest power; near a singularity it is the factor that blows up. Everything else is negligible there, and discarding it is how the yardstick is chosen.
Limit comparison replaces an inequality by a limit of the ratio $f/g$. It is the right tool when $f$ is comparable to $g$ without being neatly above or below it.
Comparison test. Suppose $0 \le f(x) \le g(x)$ for all $x \ge a$. Then
And nothing else. Below a divergent integral says nothing; above a convergent one says nothing. Two of the four pairings are useless, and they are the two that get used.
Why it is true. $F(b) = \int_a^b f$ is increasing (because $f \ge 0$) and bounded above by $\int_a^\infty g$ (because $f \le g$). An increasing bounded function has a limit — that is completeness — and the limit is the integral. Non-negativity is not a technicality: without it, $\int_a^b f$ need not increase, and the argument has nothing to stand on. A test is its hypotheses. Every convergence test in this course is a theorem with conditions, and a test quoted where its conditions fail has not been applied — it has been guessed with. Naming the hypothesis you checked is part of the answer, not a flourish on top of it.
Limit comparison. If $f, g > 0$ and $$\lim_{x\to\infty}\frac{f(x)}{g(x)} = L \quad\text{with } 0 < L < \infty,$$ then the two integrals converge or diverge together. This is usually easier to apply than a direct inequality, because you never have to argue which of $f$ and $g$ is larger — and a constant factor cannot change a verdict.
Choosing the yardstick. Out at infinity, keep the dominant power on top and bottom and cancel: $\dfrac{x + 3}{x^3 - 1}$ behaves like $x^{-2}$. Near a singularity at $c$, keep the factor that vanishes: $\dfrac{1}{\sqrt{x}\,(1+x)}$ behaves like $x^{-1/2}$ near $0$.
Only the end matters. Convergence at infinity depends on the tail alone, so an inequality that holds only for $x \ge N$ is enough — the finite part $\int_a^N$ is a proper integral and contributes a finite amount whatever it is.
Another way: picture
Two curves, $f$ underneath $g$, both heading to zero. If the area under the upper one is finite then the area under the lower one is trapped and must be finite too. Now reverse it: if the area under the lower one is infinite, the upper one has no escape either. Those are the two statements the test makes, and drawing them is the fastest way to remember which pairings are useless.
Another way: steps
Suppose $0 \le f \le g$ and $\int g$ diverges. What follows about $f$? Nothing, and two examples show why.
Take $g = \dfrac1x$ on $[1,\infty)$, which diverges. Then $f = \dfrac{1}{x^2}$ satisfies $0 \le f \le g$ and converges, while $f = \dfrac{1}{2x}$ also satisfies $0 \le f \le g$ and diverges. Both are legitimate, so the hypothesis cannot decide between them.
The same construction runs the other way: above a convergent integral, $f$ may converge (being $2g$) or diverge (being $g + \frac1x$).
The rule that survives is one sentence: a comparison is only useful when it pushes towards the conclusion you want. To prove convergence, get underneath something convergent. To prove divergence, get above something divergent. If your inequality points the other way, it is not a weaker argument — it is not an argument.
This is also why limit comparison is worth having. It is symmetric in $f$ and $g$: once the ratio has a finite non-zero limit, either integral's verdict transfers to the other, and you never have to get the direction right.
Comparing without checking non-negativity. The test is stated for non-negative functions and is false without it. $f(x) = \frac{\sin x}{x}$ needs a different argument.
Using a useless direction. See above: below divergent, or above convergent, settles nothing.
Comparing at the wrong end. $\dfrac{1}{\sqrt x (1+x)}$ behaves like $x^{-1/2}$ near $0$ and like $x^{-3/2}$ at infinity. One comparison per end.
A limit comparison with $L = 0$ or $L = \infty$. Those cases do give one-directional information, but not the two-way equivalence, and quoting the equivalence anyway is a real error.
Insisting the inequality hold everywhere. For convergence at infinity it need only hold eventually.
Reporting the ratio's limit as the integral's value. It is neither, and it never was.
Forgetting to state the yardstick's verdict. "It behaves like $x^{-1/2}$" is half an argument; the other half is that $\int_1^\infty x^{-1/2}$ diverges.
"It looks like $1/x^2$" is not a proof, and it is what most attempts at these problems amount to. The test is a logical implication with a direction, and the direction is the entire content.
The shape of the error is always the same. A learner writes an inequality that is true — $\dfrac{1}{x^2 - 1} \ge \dfrac{1}{x^2}$, say — notices that the yardstick converges, and concludes convergence. But being above a convergent integral is one of the two useless pairings; the inequality is correct and the argument is empty. The fix here is to compare with $\dfrac{2}{x^2}$ instead, which does trap the integrand from above for $x \ge 2$.
So the discipline is to write down, every time, three things: the inequality, which way it points, and the yardstick's verdict. If the sentence "$f$ is below something convergent" or "$f$ is above something divergent" cannot be said, there is no argument yet — only a resemblance.
Limit comparison exists largely to remove this trap. Once the ratio has a finite non-zero limit the two integrals are equivalent in both directions, so no direction has to be chosen and the verdict transfers either way. When a direct inequality is awkward to establish, that is the tool to reach for.
$\displaystyle\int_1^\infty e^{-x^2}dx$: there is no antiderivative, so evaluation is not available.
The situation the test exists for.
For $x \ge 1$ we have $x^2 \ge x$, so $e^{-x^2} \le e^{-x}$, and both are positive.
The inequality holds on the range, and is argued rather than asserted.
$\displaystyle\int_1^\infty e^{-x}dx = e^{-1}$ converges, and our integrand is underneath it, so the integral converges — and moreover is at most $e^{-1} \approx 0.368$. The true value is about $0.139$, so the bound is honest if not tight.
Below a convergent yardstick.
$\displaystyle\int_1^\infty\frac{dx}{\sqrt{x^3 + 2x}}$: for large $x$ the integrand behaves like $x^{-3/2}$, but is it above or below? Not obvious.
Choose the yardstick by dominant terms.
Take $g = x^{-3/2}$ and form the ratio: $\dfrac{f}{g} = \dfrac{x^{3/2}}{\sqrt{x^3 + 2x}} = \dfrac{1}{\sqrt{1 + 2x^{-2}}} \to 1$.
Finite and non-zero, so the two behave alike.
$\int_1^\infty x^{-3/2}dx$ converges, since $\tfrac32 > 1$, so the given integral converges. The inequality was never needed, and the direction never had to be decided.
Dominant terms: $\dfrac{x}{x^{2}} = \dfrac{1}{x}$. So the yardstick is $\displaystyle\int_1^\infty\frac{dx}{x}$, which diverges.
Keep the highest power above and below.
Limit comparison: $\dfrac{f}{g} = \dfrac{x(x+1)}{x^{2}+3} = \dfrac{1 + x^{-1}}{1 + 3x^{-2}} \to 1$, finite and non-zero.
The ratio settles it without any inequality.
So the two do the same thing, and the yardstick diverges: the integral diverges. Note that the integrand does tend to zero — and that this, as always, proves nothing.
Does this settle whether $\int_1^\infty f$ converges: $0 \le f \le g$ on $[1, \infty)$ and $\int_1^\infty g$ converges?
Build the proof that if $0 \le f \le g$ for every $x \ge 4$ and $\int_{4}^{\infty}g$ converges, then $\int_{4}^{\infty}f$ converges.
This task has no paper form; do it on a device.
Match each integrand to the $p$-integral it should be compared with out at infinity.
| $\dfrac{1}{x^{2}}$ | $\dfrac{1}{\sqrt{x}}$ | $\dfrac{1}{x^{3}}$ | |
|---|---|---|---|
| $\dfrac{1}{x^{2} + 3}$ | |||
| $\dfrac{x}{x^{3} + 3}$ | |||
| $\dfrac{1}{\sqrt{x} + 3}$ | |||
| $\dfrac{x^{2} + 3}{x^{5}}$ |
For each integral over $[1, \infty)$, give the exponent of the $p$-integral it behaves like, and the verdict.
| Exponent | Verdict | |
|---|---|---|
| $\displaystyle\int_1^\infty\frac{dx}{x^{3} + 3}$ | ||
| $\displaystyle\int_1^\infty\frac{x\,dx}{x^{2} + 3}$ | ||
| $\displaystyle\int_1^\infty\frac{dx}{x\sqrt{x} + 3}$ |
To test $\displaystyle\int_1^{\infty}\frac{5x^{2} + 1}{6x^{4} + x}\,dx$ by limit comparison with $\displaystyle\int_1^\infty\frac{dx}{x^{2}}$, find $\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)}$.
Answer:
$\displaystyle\int_{5}^{\infty}\frac{dx}{x^{2} + x^{4}}$ cannot be evaluated in your head, but it can be bounded. Give the best bound obtainable by comparing with $\dfrac{1}{x^{2}}$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Does this settle whether $\int_1^\infty f$ converges: $0 \le f \le g$ on $[1, \infty)$ and $\int_1^\infty g$ converges?
You can decide whether an improper integral converges without evaluating it, and say which direction your comparison points. Say in your own words why being below a divergent integral settles nothing. Next: sequences, where the same questions are asked of a list of numbers.
10. Your turn: does $\displaystyle\int_1^{\infty}\frac{x + 1}{x^{2} + 3}\,dx$ converge?, step 3