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 $0<\epsilon<1.$

**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**

*Proof.*$$\frac{[2k\pi-\theta]}{[2k\pi+\theta]}<1,$$

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).