I realize this is a very specific question, but it is something I have noticed over the last few semesters teaching sequences and series in Calculus 2 to undergraduates.

The purpose of the Limit Comparison Test is to take an intuition, such as $$\text{The series }\sum_{n=1}^\infty \frac{5\sqrt{n}-2}{n+1}\text{ ''behaves like'' }\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\text{ , which we know is divergent}$$ and make it more rigorous. I take great care in explaining how the limit $$\lim_{n\to\infty} \frac{(5\sqrt{n}-2)/(n+1)}{1/\sqrt{n}} = \lim_{n\to\infty} \frac{5n-2\sqrt{n}}{n+1} = \lim_{n\to\infty}\left(\frac{5n}{n+1} - \frac{2\sqrt{n}}{n+1}\right) = 5-0 = 5$$ yielding a finite, nonzero result shows that the terms of the two series approach 0 at rates that are approaching constant multiples of each other, thereby showing that their convergent/divergent behavior is identical. We go through several examples, each time saying, "Well, it looks like this series is 'kinda like' $\sum 1/n^2$, so let's use Limit Comparison to test it!"

This is how the test is presented in every calculus textbook I have seen, and how I have taught it in the past. I try to give more of an explanation as to how the test works, but I definitely point out how it needs to be used to test our intuitions.

Perennially, though, I notice at least a few (and sometimes shockingly many) students "inventing" their own form of the Limit Comparison test and using it on homework and exams. It goes something like this: $$ \lim_{n\to\infty} \frac{5\sqrt{n}-2}{n+1} = \text{(by l'Hôpital)} \lim_{n\to\infty} \frac{5/(2\sqrt{n})}{1} = \lim_{n\to\infty} \frac52 \cdot\frac{1}{\sqrt{n}} \text{ and we know $\sum \frac{1}{\sqrt{n}}$ diverges}$$

Quite often, this method is not even presented so nicely, showing something more like this: $$\frac{5\sqrt{n}-2}{n+1} \rightarrow \frac{5/(2\sqrt{n})}{1}\rightarrow \frac52 \cdot\frac{1}{\sqrt{n}} \rightarrow \text{diverges}$$

I have a few questions about this phenomenon:

  1. Have you also noticed this before? How did you react?
  2. Why does this occur? Are there texts/courses/teachers out there who teach this as a valid method? If not, are students tapping into some common ideas that make them come up with this (seemingly) independently every semester?
  3. (Whether or not this is currently being taught) are there good reasons to do so?
  4. If this is not a good approach to use, how can we show this to students? It does seem to be "effective" somehow, so how might we dissuade them? Should we show them outright and say, "Don't do this"? Or do we handle it only with those who re-invent this particular wheel?
  • $\begingroup$ I don't think I've seen that. I mostly remember invention of the $a_n \rightarrow 0$ thus $\sum a_n$ converges rule. Always popular, usually wrong. $\endgroup$ Commented Apr 19, 2014 at 20:40
  • $\begingroup$ @james cook: Yes, I notice that quite often, too. But that's a different issue. $\endgroup$ Commented Apr 19, 2014 at 20:42
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    $\begingroup$ Seems like a reasonable method, right? I mean, rather than following an intuition that the series behaves like some other series, you can just evaluate the limit and conclude that it diverges. $\endgroup$
    – Ruben
    Commented Apr 19, 2014 at 21:51
  • 1
    $\begingroup$ Yes, it's very reasonable. So is it taught this way elsewhere, and I'm just unaware? $\endgroup$ Commented Apr 19, 2014 at 21:53
  • $\begingroup$ Looks fine to me. They are using a known result (l'Hôpital) to compute a limit they want. $\endgroup$
    – vonbrand
    Commented Apr 20, 2014 at 3:29

1 Answer 1


I have also observed students doing this, and then I gave a 'proof' of it, and told a friend I was going to publish it. Then I did a literature search and discovered that it is false. Even though the limit before and after L'Hopitals must be the same, the rate of convergence can change. See this question and this question on Math Stackexchange, where they give examples and counterexamples.

I believe that this approach comes from a fundamental misunderstanding of the difference between sequences and series.

I believe that you should emphasize that L'Hopitals rule has to go from start to finish, i.e. it can't be used to change one function into another unless you plan on taking the limit and getting a number at the end. Once you get that number, the intermediate steps cannot be used for an other purpose.

I would use the linked examples (especially the first) to show them it doesn't work; although some students may have difficulty with the counterexample, in which case you can use a CAS to 'prove' that the counterexample doesn't work.

  • $\begingroup$ thank you Brian, I like your answer way better than mine, intuitively I thought it to be true as well, you saved me from running out and publishing my 'proof' :-) $\endgroup$
    – user1040
    Commented Apr 21, 2014 at 0:03
  • $\begingroup$ I showed my class this thread and this example, today. I think it made an impression :-) $\endgroup$ Commented Apr 25, 2014 at 17:52

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