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How should I teach Comparison Tests in Calculus II, and why?

Note that I will cover comparison tests in some way, and students will be expected to justify their answers to questions about series convergence by using tests, but they will not be expected to reproduce or even understand the proofs of the tests. Within these constraints, what do you think, or what do you do, and why?

Below are some options for "how" to get the discussion started, but I'm most interested in your "why" in any case.

  1. Teach the Limit Comparison Test only, and avoid any problems that require Direct Comparison.

    Even some oddballs like $\sum \frac{\ln n}{n}$ yield to the integral test, and we can avoid asking students about anything that would require a direct comparison, like $\sum \frac{5 + \sin n}{n}$.

  2. Teach both the Limit Comparison Test and the Direct Comparison Test, and strategies for choosing.

    Most things are handled with limit comparison, and we can handle oddballs like $\sum \frac{5 + \sin n}{n}$ with direct comparison. We teach both methods and emphasize that the students should use the "Easy" method in general and the "Hard" method for harder problems.

  3. Teach the Direct Comparison Test only; there is no reason to introduce Limit Comparison.

    Everywhere limit comparison is useful, direct comparison would have worked too, so we can simplify our course by omitting the "easy" test and focusing on getting students to be more expert at the estimates needed for Direct Comparison.

I've used textbooks that use approaches (2) and (3) and I also have had colleagues who would prefer (1), and there are probably other options, so any insight you all have would be nice.

Thanks! See also how to make Calculus II seem motivated, interesting, and useful.

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  • $\begingroup$ I think this is the key issue to examine to guide your answer: "Note that at the level I am teaching, I must cover comparison tests in some way..." Probably most of the kids don't need extensive near term use of convergence tests (not needed for calc based physics). For the very few who go on the advanced calculus, they can get more exposure later. Also, this stuff tends to come at the end of the year. And I don't know if your kids are crushing the normal parts of the course. So not an exact 1/2/3 answer, but I would lean to "satisfy the need to touch it, touch it, move on". $\endgroup$ – guest Apr 4 '18 at 16:15
  • $\begingroup$ I have to teach both, in Calculus I, and then several variables, all in one semester :( Yeah, it's crazy. $\endgroup$ – Miguel Apr 4 '18 at 17:06
  • $\begingroup$ To be clear, I think there is value in comparison tests as the right way to think about series. So please assume I plan to spend ~2-3 hours of class time on the comparison test or tests that I choose, and do not try to answer the question by saying something like "don't spend any time on this topic." I will edit "I must cover" to "I will cover" since I accidentally implied that I don't want to cover them. $\endgroup$ – Chris Cunningham Apr 4 '18 at 20:35
  • $\begingroup$ Chris, all I am saying is a prioritization comment. If your kids are killing it on the "regular calculus" than spend less time on that and more time on series (or even baby diffyQs). If they are weak on max/min problems, than do the absolute minimum on series and get them up to speed on max/min. $\endgroup$ – guest Apr 4 '18 at 20:50
  • $\begingroup$ @ChrisCunningham I will delete my comment now. I had a few minutes today so I expanded my thought a bit. $\endgroup$ – James S. Cook Apr 7 '18 at 18:01
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Another big advantage of direct comparison (if you have the time) is that it can be used to establish error estimates.

For instance, if I want to estimate

$$ \sum_1^\infty \frac{n+1}{n^4+n+7} $$

to within $\frac{1}{100}$.

Then I can make the estimates

$$ \frac{n+1}{n^4+n+7} \leq \frac{2n}{n^4} = \frac{2}{n^3} $$

Now, by drawing a picture we can see that

$$ \begin{align*} \sum_N^\infty \frac{2}{n^3} &< \int_{N-1}^\infty \frac{2}{x^3} \textrm{ d} x\\ &=\lim_{b \to \infty} \left. \frac{-1}{x^2} \right\vert_{N-1}^{b}\\ &=\frac{1}{(N-1)^2} \end{align*} $$

So we can ensure that the tail of our series is less than $\frac{1}{100}$ if we can make

$$ \frac{1}{(N-1)^2} < \frac{1}{100} $$

This first occurs when $N=12$.

Since almost every series usually covered in Calc 2 can be analyzed by comparison with a geometric series or a p-series, one can always either use the explicit sum of a geometric series, or a p-integral, to estimate the tail of the series. So you can always decide how many terms you need to obtain a given level of accuracy.

The limit comparison test gives no such explicit quantitative information: only the bare fact of convergence or divergence.

These kinds of questions force students to really think through what the tests are saying. They also give students a reason to care about them: in an application if we obtain a series as a solution to a problem, we do not only need to know if it converges or diverges, we have to estimate the sum to a desired level of accuracy.

I had good success getting students to think through this during a summer calculus session. Several students told me that the experience changed their perspective on mathematics!

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  • $\begingroup$ This is an excellent point and I am always moving more and more toward a bigger focus on error estimates. Cool! $\endgroup$ – Chris Cunningham Apr 6 '18 at 3:04
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I believe option #2 (teach both and methods for choosing) is best. I'll try to illustrate with examples.

  1. $\displaystyle{\sum_{n=1}^\infty \frac{n}{n^2+1}}$ The intuition (which most students see quickly) is that this series is "pretty much" the same as $\sum \frac{1}{n}$ because the terms being added approach $\frac{1}{n}$ "in the long run" (as $n\to\infty$). However, direct comparison does not work here (at least not easily), because the inequality $\frac{n}{n^2+1}<\frac{n}{n^2}=\frac{1}{n}$, which means the series in question is "less than $+\infty$" which tells us nothing meaningful. This is precisely where the Limit Comparison Test is helpful: when our intuition says, "This series is pretty much like another series", but the Direct Comparison is not fruitful.
  2. $\displaystyle{\sum_{n=1}^\infty \frac{n}{n^3+1}}$ Again, the intuition is that this series is "pretty much" $\sum\frac{1}{n^2}$. But in this case, the Direct Comparison test works: $\frac{n}{n^3+1}<\frac{n}{n^3}=\frac{1}{n^2}$. Thus, the series in question is "less than a finite value" which means it is also finite, and therefore convergent. Certainly, the Limit Comparison Test works here, as well, but I believe students will agree that the Direct Comparison is far more (ahem) direct and efficient. Moreover, it relies on (and reminds students of) the distinction between convergent ("adds to a finite value") and divergent.

My example #1 is, I believe, a reason to teach Limit Comparison. There are, indeed, examples (like this one) where inequalities do not work out the way we wish and Limit Comparison can be used to confirm our intuition. Furthermore, I think this particular intuition ("the terms are pretty much $1/n$") is obvious to almost all students fairly quickly. Showing them that there is a test to confirm that intuition will (a) appeal to students who intend on studying further math, especially analysis, and want to know why intuitions are valid, and (b) at least help the other students (who just want to know "what the answer is") solidify their intuition and give them practice finding an answer.

My example #2 is, I believe, a reason to teach Direct Comparison. There are, indeed, examples (like this one) where inequalities do work out the way we wish and a Direct Comparison is clear, succinct, and effective. This should also help students solidify their understanding of concepts regarding con/divergence.

Furthermore, it should be especially helpful with students who intend to pursue further math, especially analysis. Just the other day, I was working with students in Real Analysis and I appealed to precisely this example! They sought to prove $\lim_{n\to\infty} \frac{n^2}{n^4+1}=0$ using the $\varepsilon$ definition of a sequential limit. They were struggling with the algebraic steps in the inequalities. I asked them, "Could we establish an intermediary goal, one that is (a) easier to prove and (b) would imply what we need?" Using this exact example, and appealing to their intuition from Calculus II, I was able to guide them into realizing that $\frac{n^2}{n^4+1}< \frac{1}{n^2}$ and that it is much easier to prove $\frac{1}{n^2}\to 0$ using the formal definition.

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    $\begingroup$ I also (must, and do) teach both Direct Comparison and Limit Comparison, and I use examples like your #2 first to show that Comparison can work nicely. Direct comparison hardly requires a proof for students to understand. We then look at examples like your #1, showing that comparison can still be done with some imagination: $$\frac{n}{n^2+1} \geq \frac{n}{n^2+n^2} \geq \frac{1}{2n}$$ Students then ask "Do we have to come up with something like that?" Of course not, so we now have a need for a new test. $\endgroup$ – Nick C Apr 4 '18 at 22:57
  • $\begingroup$ Some students like to try to use Direct Comparison, perhaps because it feels like a challenge, or (potentially more likely) because it doesn't require any calculus. Either way, I think teaching Direct Comparison is a good and reasonable first test to teach (after, perhaps, examining $\lim_{x\rightarrow\infty}a_n$) since it requires them to exercise some number sense with functions. Eventually I tell them it's up to them to utilize some appropriate tool for the job, and preferences develop. $\endgroup$ – Nick C Apr 4 '18 at 23:05
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    $\begingroup$ @NickC 's example looks clever, but actually it can be taught as a 100-level "method" -- every "unimportant term" can always be either "discarded" or "upgraded", and one of the two always works with the inequality direction you want, so Direct Comparison is always successful. This is how the books work that choose my option (3) above. Hm. Maybe this comment should be expanded into an answer to my own question. I'd prefer to get others' answers first, I think. $\endgroup$ – Chris Cunningham Apr 5 '18 at 2:27
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You should do option (2). As you already mentioned, there are series that cannot be done through limit comparison and can only be done with direct comparison. Why would you purposely leave those out? (OK, I can think of a few reasons, but they all lead me down the rabbit hole that is "What's the point of Calculus II?") Anyway, from a mathematical point of view, this is sufficient to prove to me that both methods are necessary.

From a pedagogical point of view, having the two options (direct and limit) forces the student to actually think. If you only teach Limit Comparison, I often feel like they're turning a crank on a machine they really don't understand and/or appreciate. If you only teach Direct Comparison, there's a lot of wasted time trying to find just the right clever manipulation to make it all work. When you have both, the student is challenged to think: Can I do this easily with one method vs. the other? What does "easy" mean to me?

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    $\begingroup$ +1 I think you nailed it with your second paragraph! $\endgroup$ – Dave L Renfro Apr 5 '18 at 22:16
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While the harmonic series shows us that $a_n$ tending to $0$ is not sufficient to guarantee convergence, the comparison and limit comparison tests are strong "almost replacements": they justify the important idea that a series of positive terms will converge if its terms tend to $0$ quickly enough.

For the typical examples of series appearing in a calculus course, the limit comparison test settles so many convergence questions so quickly (provided you already know the way geometric series and $p$-series work) that it would be a severe disservice to leave that topic out, almost like forcing students to use the limit definition of a derivative by withholding rules of differentiation from them. (Well, not that bad, but it would still be pretty bad.)

The limit comparison test lets you discuss the effect of growth rates on series convergence in a way the comparison test does not. Its purpose is to let you efficiently simplify the terms under consideration: lower-order additive parts of the terms in a (positive) series can be dropped. That is the message of this test. Only having the comparison test available forces the student to deal with rather tricky inequalities (unless you deliberately water things down), suggesting to the student that the key ideas involved here are much more complicated than they need to be. For the engineering and physics students in your class, practice with the idea of neglecting all lower-order terms is worthwhile.

The comparison test itself should not be ignored, because the limit comparison test alone lets you drop lower-order additive pieces of the terms in a series, but how would you deal with $\sum 1/(n2^n)$? You can't use the limit comparison test here, and trying the comparison with both $1/n$ and $1/2^n$ is a nice, well, comparison to discuss. Something else to keep in mind, even if it is beyond the scope of the course, is that the proof of the limit comparison test uses the comparison test. Also, the proof of the ratio test uses the comparison test with a geometric series; I point that out in class to indicate why the ratio test criteria sort of look like the geometric series convergence criteria. (This is also why it would be circular to use the ratio test to explain convergence of geometric series, like the circularity of using L'Hospital's rule to find the limit of $(\sin x)/x$ as $x \rightarrow 0$.) Concerning my example above of $\sum 1/(n2^n)$, it of course can be handled "without thinking" by using the ratio test, and doing this example first by the comparison test and later by the ratio test allows a sense of progress to better (or at least a wider range of) tools to be displayed to the class.

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    $\begingroup$ My calculus book has cases for limit comparison where the limit is $0$ or $\infty$. So limit comparison technically applies to your example (at least for some conventions). $\endgroup$ – Steven Gubkin Apr 9 '18 at 15:37
  • $\begingroup$ Okay, if $a_n/b_n \rightarrow 0$ (with $a_n, b_n > 0$) then for large enough $n$ we have $a_n/b_n \leq 1$ and thus $a_n \leq b_n$ for $n \gg 0$, so by the comparison test if $\sum_{n\geq 1} b_n$ converges then so does $\sum a_n$. But this is not an equivalence of convergence for $\sum_{n \geq 1} a_n$ and $\sum_{n\geq 1} b_n$. What I teach in my class if that when two positive sequences grow (or decay) in the same way, meaning $a_n/b_n \rightarrow 1$, then whether or not $\sum_{n\geq 1} a_n$ or $\sum_{n\geq 1} b_n$ converge are equivalent issues. $\endgroup$ – KCd Apr 9 '18 at 23:16
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    $\begingroup$ Yes, I usually just teach this way as well because it is generally easier to just think through the comparison. I use the phrase "the series are holding hands". Another nitpick: do you require your students to pick $b_n$ so that the limit is 1? $\endgroup$ – Steven Gubkin Apr 10 '18 at 0:02
  • $\begingroup$ The only case I discuss is when $a_n \sim b_n$, which is fairly natural anyway. $\endgroup$ – KCd Apr 12 '18 at 0:11
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I tend to teach option (2.) for many of the reasons already listed in the existing answers. If we just teach (1.) then the problem solving becomes formulaic which means the whole endeavor is essentially pointless. Why do we study convergence and divergence in calculus II? It's not really about learning whether these series converge or diverge. The real reason is to expose students to problems which require analysis. In particular, the use of theorems, attention to estimates, and perhaps most importantly the idea that $\Rightarrow$ is not the same as $\Leftrightarrow$. For example, a student who succeeds is one who realizes that showing $a_n \rightarrow 0$ does not imply $\sum a_n $ converges despite the fact that $a_n \nrightarrow 0$ does implie $\sum a_n$ diverges. Or for that matter, I view it as a success if the student has at least understood that the convergence of $a_n$ and $\sum a_n$ are related, but distinct, questions. So, again, the point of studying convergence and divergence is to introduce them to mathematical reasoning. Admittedly, they've had doses of it here and there, but in the study of convergence and divergence we have something like one-dozen theorems to apply. It is way harder logically then their previous work. I often argue it doesn't really belong in the standard calculus sequence because it is really at a completely different level if we are to do it well. Again, I fear not doing it well makes the whole discussion worthless.

Pragmatically, what students really need to succeed in engineering and physics is a mastery of power series calculation. Excellence in applying the dozen or so convergence and divergence tests does not help me a bit when asked to find the series expansion of $f(x) = \frac{1}{1+x^2}$ or the first three terms in $\sin(e^x)$ or $\tan(2x+1)$ or you name it... Sure, convergence logically matters, but the binomial series is far more important than the limit comparison test.

So, again, if I am to teach convergence and divergence it really should be with the goal of getting students to be better at mathematical analysis. With that goal in mind the choice of (2.) is clear since it requires the most sincere analysis from students.

That said, there is another tool worth exploring which I don't think anyone has yet to cover. Substitution: consider, $$ \sum_{n=1}^{\infty} \frac{1}{n^2+6n+9} = \sum_{n=1}^{\infty} \frac{1}{(n+3)^2} $$ Set $k=n+3$ then $n=1$ provides $k=4$ and thus, $$ \sum_{n=1}^{\infty} \frac{1}{n^2+6n+9} = \sum_{k=4}^{\infty} \frac{1}{k^2}. $$ Since this is the tail of the $p=2$ series it converges (since $p=2$ converges and the tail of any convergent series likewise converges). But, if we know the $p=2$ series converges to $\pi^2/6$ then we can calculate the tail by some simple arithmetic. $$ \sum_{n=1}^{\infty} \frac{1}{n^2+6n+9} = \sum_{k=4}^{\infty} \frac{1}{k^2} = \sum_{k=1}^{\infty} \frac{1}{k^2} - 1 - \frac{1}{4}-\frac{1}{9} = \frac{\pi^2}{6} - 1 - \frac{1}{4}-\frac{1}{9}. $$ Combining these ideas with the direct comparison test leads to further options for analysis. But, sometimes, the Limit Comparison Test is just a better option if the problem is just about convergence.

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I think we should at least mention direct comparison, and show a brief example like $\sum \frac{1}{x^2+1} < \sum \frac{1}{x^2}$, because that seems like the most straightforward way for students to develop some feeling for why any comparison-type results for series should be true in the first place.

But I also think it's fine to underemphasize direct comparison after it first comes up, and point out that limit comparison is more "robust" because you can't have the phenomenon of an unlucky inequality symbol that points in the "wrong" (i.e. unhelpful) direction.

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  • $\begingroup$ Thanks for the answer! To clarify, would you say the "analysis estimate challenge" of getting the inequality symbols to go the right way in the direct comparison test are generally too complex to be worthwhile, and that would be your main reason to de-emphasize direct comparison as soon as possible? $\endgroup$ – Chris Cunningham Apr 4 '18 at 20:39
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I would lean towards 1. You need to prioritize. Your students don't have unlimited time or ability. A very sound case could be made for not including this content at all. [Note, I like the content, I did the content when I took the course, blabla. I am just saying priorities are a factor!] Given your likely situation based on school, student population, needs of the student for future course, this stuff is a bell/whistle. Do the minimum. I'm sure there is somewhere else some attention could be used!

So choice 1 over 2 as "less stuff". [You can either give some time back to the "normal calculus" or just do a less rushed coverage of the limit method if you keep time constant.] Choice 1 over choice 3 as the more important method and what people expect the kids to have covered when you say you are expected to hit the topic.

When you assign homework, obviously make sure that none of the problems need the method you don't teach and you might just mention that you are only teaching one method (without even describing the other method...just so they know they can't use the alternate problems for drill or so they know "don't need to know it for the test".) Any kids who are motivated, can obviously learn the direct method and drill it from the text. But be clear that you are limiting scope for the course.

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  • $\begingroup$ Is there a way to see who has up/down voted me? Does it depend on if I am a second class citizen or not? $\endgroup$ – guest Apr 6 '18 at 19:44
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    $\begingroup$ As far as I know no one can tell. You can see how many votes of each type appear by clicking on the score: right now it says +1, 0 so one person upvoted, and no one downvoted. $\endgroup$ – Chris Cunningham Apr 6 '18 at 21:01
  • $\begingroup$ Were you the up? $\endgroup$ – guest Apr 6 '18 at 21:02
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    $\begingroup$ As far as I know no one can tell. $\endgroup$ – Chris Cunningham Apr 7 '18 at 14:12
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I would like to give the argument for option (3). Even though I am not necessarily a strong supporter of it, my calculus book does it and I think it's fine to do so: to present the Direct Comparison Test and skip the Limit Comparison Test.

The most common argument against this is that students need to be too clever to get the right estimates, but it does not require too much cleverness if presented carefully:

  • First try to guess what the answer is by looking at "important terms."
  • Use this guess to choose either $\leq$ or $\geq$.
  • For each "unimportant" term, either downgrade the term or upgrade it to be important.
  • At the end, make sure that the comparison test gives a result.

For this question:

Find whether $\sum_{n=2}^\infty \frac{4n^2 - 5}{7n^4 - 5}$ converges or diverges using the comparison test.

A student's thought process would then be:

Guess: $\sum \frac{n^2}{n^4}$ converges because it is a p-series $p=2>1$, so we try to show it converges.

So start with $\frac{4n^2 - 5}{7n^4 - 5} \leq$ .

Can I drop the 5 on top? $\frac{4n^2 - 5}{7n^4 - 5} \leq \frac{4n^2}{7n^4-5}$ Yes, that's fine.

Can I drop the other 5? $\frac{4n^2 - 5}{7n^4 - 5} \leq \frac{4n^2}{7n^4-5} \leq \frac{4n^2}{7n^4}$ No, that's not true.

So I'll upgrade it instead: $\frac{4n^2 - 5}{7n^4 - 5} \leq \frac{4n^2}{7n^4-5} \leq \frac{4n^2}{7n^4 - 5n^4}$ which is true.

Then the student's solution is:

$\frac{4n^2 - 5}{7n^4 - 5} \leq \frac{4n^2}{7n^4-5} \leq \frac{4n^2}{7n^4 - 5n^4} = \frac{4n^2}{2n^4} = 2\frac{1}{n^2}$. The series 2 $\sum\frac{1}{n^2}$ converges because it is twice a p-series with $p>1$, and the original series is smaller, so it converges by the comparison test.

My experience is that this kind of solution is very reachable by Calc 2 students, as it has been reduced to a "process" that can be digested. It also still requires students to use their intuition to guess at the answer (to decide whether to start with $\leq$ or $\geq$) but then makes it clear that your intuition is not the entire solution.

Separating out that the student's intuition is different from the student's solution is likely the key thing we are getting out of teaching series tests at all. So this approach seems somewhat elegant to me.

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  • $\begingroup$ Disagree, particularly with the 5 to $5n^4$ part, but argument is excellent. $\endgroup$ – BCLC May 1 '18 at 16:37

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