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Dealing with a recent question I spotted a very nice exercise for Calc-2 students, i.e. to find the mistake in the following lines.

Lemma 1. For any $n\in\mathbb{N}$, we have: $$ \int_{0}^{1} x^n\left(1+(n+1)\log x\right)\,dx = 0. $$ Lemma 2. For any $x\in(0,1)$ we have: $$ \frac{1}{1-x}=\sum_{n\geq 0}x^n,\qquad \frac{\log x}{(1-x)^2}=\sum_{n\geq 0}(n+1) x^n\log(x). $$ By Lemmas 1 and 2 it follows that: $$\begin{eqnarray*}(\text{Lemma 1})\quad\;\;\color{red}{0}&=&\int_0^1 \sum_{n\geq0} x^n\left(1+(n+1)\log x\right)\,dx\\[0.2cm](\text{Lemma 2})\qquad&=&\int_0^1 \left(\frac{1}{1-x} + \frac{\log x}{(1-x)^2}\right)\,dx\\[0.2cm](x\mapsto 1-x)\qquad&=&\int_0^1 \left(\frac{1}{x}+\frac{\log(1-x)}{x^2}\right)\,dx\\[0.2cm](\text{Taylor series of }x+\log(1-x))\qquad&=&-\int_0^1 \frac{1}{x^2} \sum_{k\geq2}\frac{x^k}k \,dx\\[0.2cm](\text{termwise integration})\qquad&=&-\sum_{k\geq 2} \frac{1}{k(k-1)}\\[0.2cm](\text{telescopic series})\qquad&=&-\sum_{m\geq 1} \left(\frac{1}{m}-\frac{1}{m+1}\right)=\color{red}{-1}. \end{eqnarray*}$$

Now the actual questions: were you able to locate the fatal flaw at first sight?
Do you think it is a well-suited exercise for Calculus-2 (or Calculus-X) students?

Community
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Jack D'Aurizio
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  • I don't think this is appropriate at all for Calc-1. Possibly as an extra credit problem for an advanced class. Seems to rely a lot on knowing how to take an infinite sum--but that's not a Calc-1 topic (that would be Calc-3). I certainly do not see the flaw "at first sight". I would have to sit down and analyze how they got from each step to the next before I could say anything about where the flaw is. Perhaps it might be more appropriate with annotations for each step (but then that may blatantly expose the flaw). – Jared Jul 05 '16 at 03:29
  • I would think this is well beyond Calc I. I'm not sure what the standard is, but in my studies Calc 1 was differential calculus, and Calc 2 was integral calculus. – Carser Jul 05 '16 at 03:31
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    I also think this is way beyond the abilities of most elementary calculus students. It's a good exercise in teaching them to be skeptical about interchanging sums and integrals for a more advanced student, though, I suppose. –  Jul 05 '16 at 03:38
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    I'm almost sure that the error is in the misapplication of the first lemma. But I don't think that this is a good exercise just because I feel that if the error is that you have essentially interchanged a sum and integral then I feel that this is very subtle. Since it's always okay to interchange a finite sum and a integral, unless I was given a heads up that sometimes this is not okay with infinite sums then I would be lost by this exercise. – mike van der naald Jul 05 '16 at 03:40
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    In the U.S. calculus courses that I've taught or observed, this material would come in Calc. $2$, to the extent that it appeared at all, and the interchange of integral and summation wouldn't appear at all. That makes it an essentially impossible exercise. It might be appropriate for a good old-fashioned advanced calculus course. – Brian M. Scott Jul 05 '16 at 03:43
  • Are you tacitly treating $\displaystyle \int_0^1 \frac{dx} x$ as if it were a finite number? $\qquad$ – Michael Hardy Jul 05 '16 at 03:58
  • @MichaelHardy: no, it is more subtle. $$\sum_{n\geq 0}x^n(1+(n+1)\log x)$$ actually is pointwise convergent to $\frac{1}{1-x}+\frac{\log(1-x)}{(1-x)^2}$ on $(0,1)$, but... – Jack D'Aurizio Jul 05 '16 at 04:00
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    In the U.S. at typical state universities, vast numbers of students with no understanding of the prerequisite material are encouraged to take first-year calculus, and do so. Most of them don't suspect there's such a thing as understanding mathematics; they just think it consists of obeying their instructors on how to push symbols around. They bring in tuition money, and that's what they're for, in the view of potentates in charge. $\qquad$ – Michael Hardy Jul 05 '16 at 04:01
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    Consider post it at math educator stack exchange? –  Jul 05 '16 at 04:04
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    @MichaelHardy: the situation in Italy is quite different. We have, in fact, very good math courses and a great pressure on understanding things. The cons are that few students can actually survive that pressure, and courses are a dead end at last, since very few money is invested in University and research. – Jack D'Aurizio Jul 05 '16 at 04:05
  • $$ \int_0^1 \sup_N \left| \sum_{n=0}^N x^n (1+(n+1)\log x) \right| , dx = +\infty \text{ ?} $$ – Michael Hardy Jul 05 '16 at 04:10
  • @MichaelHardy: that is the caveat, good job! But please, to you and all the others involved, post your thoughts/opinions as answers, I will be glad to reward some reputation points. – Jack D'Aurizio Jul 05 '16 at 04:13
  • By "exercise", do you mean an exam question? –  Jul 05 '16 at 04:23
  • @AhmedHussein: it is a chance, yes. But given the respectable opinions here, I do not think I will use it for real. Or maybe just for talented students. – Jack D'Aurizio Jul 05 '16 at 04:25
  • I checked the second $\texttt{Lemma 1}$ and the result is $\color{#f00}{-1}$. OP claimed it's $\underline{zero}$ !!!. – Felix Marin Sep 16 '16 at 01:58
  • @FelixMarin: yes, the correct result is $-1$, but that is not the point of the exercise, the point is to spot the fallacy. – Jack D'Aurizio Sep 16 '16 at 12:50

2 Answers2

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The error is at the very beginning, in the interchange of integral and summation. I’m too rusty (and too lazy!) to go beyond checking that the sequence of functions fails to satisfy a standard sufficient condition for the interchange, but the other steps are legitimate, so that must be the sticking point.

In the U.S. calculus courses that I've taught or observed, this material would come in Calc. $2$, to the extent that it appeared at all, and the interchange of integral and summation wouldn't appear at all; that makes it an essentially impossible exercise. Moreover, most students in typical first-year calculus courses still have the notion that mathematics is algorithmic calculation; getting them to pay enough attention to details to understand why the non-existence of a zero of $\frac1x$ doesn’t contradict the intermediate value theorem, or even to remember that the sign of $x$ matters when multiplying an inequality $f(x)\le g(x)$ by $x$, is a non-trivial challenge, to the extent that the former often goes by the board.

This might be appropriate for a very good old-fashioned advanced calculus course; the undergraduate real analysis courses that I taught had a different emphasis and didn’t cover the necessary material.

Brian M. Scott
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$$ \int_0^1 \sup_N \left| \sum_{n=0}^N x^n (1+(n+1)\log x) \right| \, dx = +\infty $$

The dominated convergence theorem says $$ \lim_N \int f_N = \int \lim_N f_N \quad\text{if } \int \sup_N |f_N| < \infty. $$ That's too big an "if" in this case.

Michael Hardy
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