Subseries of $p$-series Let $A \subseteq \mathbb{N}$. Let $0 \leq p \leq 1$ be the unique number such that $\sum_{n \in A} n^{-p-\epsilon}$ converges and $\sum_{n \in A} n^{-p+\epsilon}$ diverges for every $\epsilon > 0$.
Can we find, for every $0 \leq q < p$, a set $B_q \subseteq A$ such that $\sum_{n \in B_q} n^{-q-\epsilon}$ converges and $\sum_{n \in B_q} n^{-q+\epsilon}$ diverges for every $\epsilon > 0$?

For any set $A \subset \mathbb{N}$, let $A(n) := A \cap \{\ell \in \mathbb{N} \colon 2^n \leq \ell < 2^{n+1}\}$. Then, for $\ell \in A(n)$ and $\alpha \in \mathbb{R}$, we have $2^{-\alpha} 2^{-\alpha n} \leq \ell^{-\alpha} \leq 2^{-n \alpha}$, and this easily implies
$$
  \sum_{\ell \in A}
    \ell^{-\alpha}
  = \sum_{n=1}^\infty \,\, \sum_{\ell \in A(n)} \ell^{-\alpha}
  \asymp \sum_{n=1}^\infty 2^{-\alpha n} \#A(n)
  .
  \tag{$\ast$}
$$
Now, define (somewhat similar to the formula for the radius of convergence of a power series)
$$
  p^* (A)
  := \limsup_{n\to\infty} \frac{\log_2 (\#A(n))}{n}
  ,
$$
noting that $0 \leq p^\ast(A) \leq 1$ (why?!).
I now claim that
$$
  \sum_{\ell \in A} \ell^{-(p^\ast(A) + \epsilon)}
  < \infty
  \quad \text{and} \quad
  \sum_{\ell \in A} \ell^{-(p^\ast(A) - \epsilon)}
  = \infty
  \qquad \forall \, \epsilon > 0
  .
  \tag{$\lozenge$}
$$

To see this, note that there exists $N_0$ such that $\frac{\log_2(\# A(n))}{n} - p^*(A) \leq \frac{\epsilon}{2}$ for all $n \geq N_0$ and thus
$$
  \log_2 \big( 2^{-(p^*(A) + \epsilon/2) n} \cdot \# A(n) \big)
  = n \cdot \Big( \frac{\log_2 (\# A(n))}{n} - p^*(A) - \epsilon/2 \Big)
  \leq 0
  ,
$$
which means that $2^{-(p^*(A) + \epsilon/2) n} \cdot \# A(n) \leq 1$ for all $n \geq N_0$. This easily implies that $\sum_{n=1}^\infty 2^{-(p^* (A) + \epsilon)n} \# A(n) < \infty$. By $(\ast)$, this implies that $\sum_{\ell \in A} \ell^{-(p^\ast(A) + \epsilon)} < \infty$, proving the first part of $(\lozenge)$.
To prove the second part, we show that actually $2^{-(p^*(A) - \epsilon) n} \# A(n) \not\to 0$, which easily implies $\sum_{n=1}^\infty 2^{-(p^*(A) - \epsilon) n} \# A(n) = \infty$, which by $(\ast)$ will imply the second part of $(\lozenge)$.
To show $2^{-(p^*(A) - \epsilon) n} \# A(n) \not\to 0$, assume towards a contradiction that $2^{-(p^*(A) - \epsilon) n} \# A(n) \to 0$. Thus, there exists $C > 0$ satisfying $2^{-(p^*(A) - \epsilon) n} \# A(n) \leq C$ and thus
$$
  \log_2 (\# A(n)) - (p^*(A) - \epsilon) n \leq \log_2 (C)
  .
$$
Hence,
$$
  \frac{\log_2 (\# A(n))}{n} \leq p^*(A) - \epsilon + \frac{\log_2(C)}{n}
  ,
$$
which is impossible by definition of $p^*(A)$. This completes the proof of $(\lozenge)$.

Finally, let $0 \leq q < p^* (A)$ be arbitrary. For each $n \in \mathbb{N}$, pick $B_n \subset A(n)$ with $\# B_n = \min \{ \# A(n), \lceil 2^{q n}\rceil \}$ and set $B = \bigcup_{n=1}^\infty B_n$. It is then not hard to see
$$
  p^*(B)
  = \limsup_{n\to\infty}
      \frac{\log_2 (\# B(n))}{n}
  = \limsup_{n\to\infty}
      \min \Big\{ \frac{\log_2 (\# A(n))}{n}, \frac{\log_2 (\lceil 2^{qn}\rceil)}{n} \Big\}
  = \min \{ p^*(A), q\}
  = q
  .
$$
By the above considerations, this easily shows that $B$ is as required.
