The series $\sum_{n\in\mathbb{N}}n^{-p}$ diverges if $p\leq1$ and converges if $p>1$, and so it may seem plausible that (being a "bifurcation point" of this condition) the harmonic series $\sum_{n\in\mathbb{N}}n^{-1}$ could converge on some proper subset of $A\subset\mathbb{N}$. This is obvious if $A$ is finite. If $A$ is infinite, then a moment's thought reveals that there are many subsets on which the harmonic series converges since its terms contain any other series with terms in $\mathbb{N}^{-1}$. So for instance $$\sum_{n\in\mathbb{N}}n^{-2}=\frac{\pi^{2}}{6},$$ $$\sum_{n\in\mathbb{N}}\frac{1}{n!}=e,$$ $$\sum_{n\in\mathbb{N}}\frac{1}{2^{n}}=2,$$ and so on. Given that rather "large" subsets of $\mathbb{N}$ lead to convergence of the harmonic series, the following result was somewhat surprising to me when I was first asked to prove it.
Claim. Let $$A_\epsilon := \{a \in \mathbb{N} : 1 - cos(a) < \epsilon\}.$$ Then $$\sum_{n\in A_\epsilon } \frac{1}{n}$$ diverges for all $0<\epsilon<1.$Proof. For $0<\epsilon< 1$, the inequality $1-\cos(a)< \epsilon$ has solutions for $$a\in(2k\pi-\theta,2k\pi+\theta)$$ where $\theta=\cos^{-1}(1-\epsilon)$ (note that $\theta\in(0,\frac{\pi}{2})$ and by using a Taylor expansion, it is easy to see $\theta=O(\epsilon^{\frac{1}{2}})$, although all that is important is $\theta\to0$ as $\epsilon\to0$). For there to be any positive integers $a:=a_{k}$ in such an interval, it is necessary and sufficient that
$$\frac{[2k\pi-\theta]}{[2k\pi+\theta]}<1,$$
where $[\cdot]$ is the "floor" function (round down, e.g. truncate the decimals). Intuitively, this condition just says there is an integer in the $k$th solution interval (note that there could be multiple integral solutions in a $k$th interval, though this is not very important since we are mostly interested in the case for small $\epsilon$; furthermore, since $\theta=O(\epsilon^{\frac{1}{2}})$, then once $\epsilon$ is sufficiently small (say $\epsilon<0.1,$ so that $2\theta$).
From the above observations and the fact that $2\pi<6.3$ (circumference of the unit circle), it is not difficult to ascertain that $\#A=\infty$ (the cardinality of the set $A$). Therefore $A$ is countable with its elements forming an "approximate" arithmetic sequence of integers in that sense that for
$$D:=\max_{a_{i}\in A}|a_{i+1}-a_{i}|<\infty,$$
on "average" the difference of two successive integers is approximately $D$ (having analytical results that are sharp is unnecessary in the present situation as we are only after qualitative facts like convergence).
We can now determine whether or not the sum converges. Define sequences $a_{j}:=\frac{1}{j}$ for $j\in A $, and $0$ otherwise, and $b_{j}:=\frac{1}{j}$ for all $j=1,2,\ldots.$ Then $c_{j}:=\frac{a_{j}}{b_{j}}=1$ for $j\in A$, and $0$ otherwise. Therefore, $c_{j}$ has a sum which looks like $$1+0+\ldots+0+1+0+\ldots+0+1+\ldots$$ Define one more sequence $d_{j}:=1$ if $j=a_{1}D$ ($a_{1}$ being the first integral solution to the original inequality) and $0$ otherwise (in other words, $d_{j}$ really is an arithmetic sequence with common difference $D$). Recall from the theory of Cesaro summation that for zero-spacing $D$, $$\frac{1+0+\ldots+0+1+0+\ldots+0+\ldots+0+1_{n}}{n}\to\frac{1}{D+1}\;as\;n\to\infty$$ (note because Cesaro summation is an averaging process, the limit holds even if there is a finite number of instances of improper spacing for a finite number of terms). Consequently, $$\frac{d_{1}+\ldots+d_{j}}{n}=\frac{1}{D+1}\;as\;n\to\infty$$ (see previous parenthetical remark). Consequently, \begin{align*} \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{j=1}^{n}\frac{a_{j}}{b_{j}} &=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{j=1}^{n}c_{j}\\ &\geq\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{j=1}^{n}d_{j}\\ &=\frac{1}{D+1}\\ &>0 \end{align*} for all $\epsilon>0$, no matter how small (note that $D$ behaves something like $O(\theta^{-1})$, and by extension something like $O(\epsilon^{-\frac{1}{2}}).$ It follows that $$\sum\limits_{j=1}^{\infty}a_{j}=\infty,$$ e.g. diverges for every $\epsilon>0$ (if you don't see why or don't recognize the convergence theorem used, just apply the summation by parts formula to $\sum a_{j}$ together with the established bound).
