My motivation for this enquiry is due to interest on Erdos conjecture on arithmetic progression. (https://en.wikipedia.org/wiki/Erd%C5%91s_conjecture_on_arithmetic_progressions)
Having come across this conjecture once again I decided I would give it a shot to at least try to understand what the conjecture states.
I have made a rough attempt to do so.
Could you please verify if my line of reasoning should be worked further or this approach is total nonsense?
Let A be a set of positive integers:$$\sum_{n\in A}\frac1n = \infty$$ Then A contains arbitrarily long Arithmetic Progressions (AP)
I state that $$\sum_{n}\frac1n\to \infty\iff \lim_{n\to\infty}\sum\frac1n$$ If n has upper bound:$$n\leqslant M, \sum\frac1{n}$$is a small set, thus converges.
$$\begin{align} \text {Define function}: \ f(x)=\frac1x\space,(x \gt 0)\space \text{and} \space a_n=\frac1n,n\in A \end{align}$$ Fig.1
Since intergral can be interpreted as area under the curve, it is clear that: $$\text S_n = \sum_{k=1}^n\frac1k \geqslant \int_{1}^{n+1}f(x)dx$$
Also $$\lim_{n\to\infty}\int_{1}^{n+1}f(x)dx = \lim_{n\to\infty}\Big[\ln\lvert x\rvert\Big]_1^{n+1} = \lim_{n\to\infty}\ln(n+1)\to\infty$$ Thus $$\lim_{n\to\infty} \sum_{k=1}^n\frac1k\to\infty$$
Integral can be also written as left Riemann sum: $$\int_1^nf(x)dx\approx \sum_{i=0}^kf(x_i)\Delta{x} \space{where}\space \Delta{x}=\frac{n-1}k\space and\space x_i=1+\Delta{x}(i)$$
For$$(n\geqslant2)$$$$\text{Let}\space\Delta{x}=1\space \Rightarrow \sum_{i=0}^{n-1}f(x_i)\Delta{x}=\sum_{k=1}^n\frac1k$$ As $$\Delta{x}\to 0$$ $$\begin{align} \int_1^nf(x)dx=\lim_{k\to\infty}\sum_{i=0}^kf(x_i)\Delta{x}\leqslant\lim_{k\to\infty}\sum_{i=0}^kf(x_i)(1) \end{align}$$
Since $$\lim_{n\to\infty}\int_1^nf(x)dx\to\infty,\space\lim_{n\to\infty}\Big[\lim_{k\to\infty}\sum_{i=0}^kf(x_i)(1)\Big]\space\text{must diverge too.}$$
$$\text{Notice that}\space x_i\space\text{forms arithmetic progressions}\space[x_i=x_0 + \Delta{x}(i)]\space and \space n\geqslant \frac1n.$$ Finally, integral can be thought as multiple addition of inverse AP terms times infinitesimal del x. Since integral diverges, airthmetic progression must diverge too, thus become arbitrarily long.