What is wrong with this proof? (divergence of the harmonic series)

[BWV]

Reading this piece with a number of proofs of the divergence of the harmonic series
http://scipp.ucsc.edu/~haber/archives/physics116A10/harmapa.pdf

and this example states: 'While not completely rigorous, this proof is thought-provoking nonetheless. It may provide a good exercise for students to find possible flaws in the argument.'

not being any good at proofs, curious what flaws or lack of rigor there is here

 

[FactChecker]

IMHO, to be rigorous, the first equality between the summation and an integral is a problem. You should not say they are equal when one (or both) are infinite. In a formal proof, you probably should show that the summation is greater than any positive number, M. You can do that by comparing the summation to the integral, but modify it to allow it to be as large as you need.

 

[PeroK]

Reading this piece with a number of proofs of the divergence of the harmonic series
http://scipp.ucsc.edu/~haber/archives/physics116A10/harmapa.pdf

and this example states: 'While not completely rigorous, this proof is thought-provoking nonetheless. It may provide a good exercise for students to find possible flaws in the argument.'

View attachment 239116

not being any good at proofs, curious what flaws or lack of rigor there is here

If you compare the partial sums in all cases, the equality should hold up well.

 

[fresh_42]

IMHO, to be rigorous, the first equality between the summation and an integral is a problem. You should not say they are equal when one (or both) are infinite. In a formal proof, you probably should show that the summation is greater than any positive number, M. You can do that by comparing the summation to the integral, but modify it to allow it to be as large as you need.

If ##(f_k)_{k\in \mathbb{N}}## is a sequence of measurable functions ##f_k\, : \,\Omega \longrightarrow \mathbb{C}## then
$$
\int_\Omega \sum_{k=0}^\infty |f_k| = \sum_{k=0}^\infty \int_\Omega |f_k|
$$
In our case, we have ##\Omega = [0,1]\; , \;f_k(x)=|f_k(x)|=x^k## which is measurable on ##[0,1]##, so
$$
\sum_{k\in \mathbb{N}} k^{-1} =\sum_{k\in \mathbb{N}_0} \int_{[0,1]} x^k \,dx= \int_{[0,1]} \sum_{k=0}^\infty x^k \,dx = \int_{[0,1]} \dfrac{1}{1-x} \,dx
$$
and the proof is valid. The crucial point is, that the set ##\{\,x\in [0,1]\,|\,\sum_{\mathbb{N}_0} |f_k(x)| = \infty \,\} = \{\,1\,\}## is of measure zero. So if we read the integral as Lebesgue integral we're fine.

 

[BWV]

Cool, thanks for the responses

 

[mathman]

Things are a little hairy at the end point of the integral. ##\sum_{k=0}^\infty x^k=\frac{1}{1-x}## for ##|x|\lt 1##. You need to handle the upper limit carefully.

 

[WWGD]

Monotone Convergence theorem (adapted to this problem) says that if/since {##\Sigma x^n ##} is a pointwise increasing sequence of measurable (since continuous; partial sums are continuous in x) functions :

##Lim_{n \rightarrow \infty}\int \Sigma x^n = \int Lim_{n \rightarrow \infty} \Sigma x^n=
\int \frac {1}{1-x} ##