# A very tricky pseudo-proof of $0=-1$ through series and integrals

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?

• I am not familiar with both the content and the prerequisites of a "Calculus-2" class. Could you give a brief outline of what the students should learn and what they should know already? – Christian Jul 5 '16 at 9:04
• Not suitable if students do not know "What conditions make $\int f_n\rightarrow \int f$?" As far as I know that question is not answered in most (lower division) calculus class I took/taught/knew. – user2139 Jul 5 '16 at 9:09
• Seems like a good question for a real analysis course rather than a standard calculus course. – John Coleman Jul 5 '16 at 15:45