Conditional convergence of $\sum_k \frac{\epsilon_k}{k}$

Question

Consider the infinite series $$ \sum_{k=1}^\infty \frac{\epsilon_k}{k}, \quad \epsilon_k = \begin{cases} +1 \quad (n = 1 \; \text{or} \; n \; \text{prime})\\ -1 \quad \text{otherwise} \end{cases} $$ The series is clearly not absolutely convergent, and I suspect that it diverges. I am interested in proofs that settle the convergence question, from elementary to advanced.

This question came up in my Calculus class today.

Answer 1

You could use Kronecker's Lemma in the form that if the sequence $\sum_{k=1}^n {\epsilon_k\over k}$ converges, then ${1\over n}\sum_{k=1}^n \epsilon_k\to 0$. This would imply, for the prime counting function $\pi(n)$, that ${\pi(n)\over n}\to {1\over 2}.$

But only two of every six consecutive integers can possibly be prime, so ${\pi(6n)\over 6n}\leq {1\over 3}$, which gives a contradiction.

Answer 2

Through Euler's product for the $\zeta$ function $$ \zeta(s) = \sum_{n\geq 1}\frac{1}{n^s} = \prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^s}\right)^{-1}\quad (s>1)\tag{1}$$ and logarithmic differentiation/summation by parts it can be proved that $$ \sum_{n=1}^{N}\frac{1}{n}\approx \log N,\qquad \sum_{p\leq N}\frac{1}{p}\approx \log\log N \tag{2} $$ hence the partial sums of your series diverge like $\log\log N-2\log N\to -\infty$.
Informally speaking, primes are not dense enough to counteract the divergence of the harmonic series. We may also directly apply summation by parts and exploit some weak form of the PNT to be able to state $$ \sum_{k=1}^{N}\epsilon_k \leq -N+\frac{C N}{\log(N)}\quad\Longrightarrow\quad \sum_{k\geq 1}\frac{\epsilon_k}{k}\quad\text{diverges.}\tag{3} $$ As suggested in the comments (thanks to user1952009), the very crude bound $\pi(N)\leq\frac{N}{3}$ (for any $N$ large enough) is already enough to prove the divergence of $\sum_{k\geq 1}\frac{\epsilon_k}{k}$ by summation by parts.