What numbers can be expressed as a subseries of a p-series? For a given real number $p>1$ we define the following set of non-negative real numbers
$$Z_p:=\left\{\sum_{n=1}^\infty\frac{\epsilon_n}{n^p}:\epsilon_n\in\{0,1\}\right\}\subseteq[0,\zeta(p)]$$
Notice that $Z_p$ is the set of subseries of the well-known $p$-series $\zeta(p)$. My question then is

Q:  Precisely which real numbers conform the set $Z_p$ for $p>1$?

I believe that I have been able to study the case $p=2$. But there probably is a much simpler proof.
$\color{green}{\bf{Proposition.}}$ For $p=2$, we have $Z_2=\left[0,\zeta(2)-1\right]\cup[1,\zeta(2)]$.
$\color{seagreen}{Proof.}$ Let $x\in[0,\zeta(2)-1]\cup[1,\zeta(2)]$. We will now define a sequence $(x_n)_{n\geq1}$ such that $x=\lim_n x_n\in Z_2$. First, define $x_1=\lfloor x_1\rfloor\in\{0,1\}$ and choose $\epsilon_1=x_1$. Then, we define recursively
$$x_n:=\begin{cases}x_{n-1} & \text{ if } x < x_{n-1}+1/n^2\\ x_{n-1}+1/n^2 & \text{otherwise}\end{cases}\quad\text{and choose}\quad\epsilon_n:=\begin{cases}0 & \text{ if } x < x_{n-1}+1/n^2\\ 1 & \text{otherwise}\end{cases}$$
It is easy to see that $\forall n\geq1:x_n\leq x$ and $(x_n)_{n\geq1}$ is monotonically increasing which means that it converges to a number $\leq x$. In fact, $\lim_n x_n\in Z_2$. We are now going to prove that $\forall n\geq1:x-x_n\leq\sum\limits_{k\geq n+1}\frac{1}{k^2}$ by induction where the base case is clear. Supose it is true for $n-1$. Then, if $\epsilon_n=1$,
$$x-x_{n}=(x-x_{n-1})+(x_{n-1}-x_n)\leq\left(\sum_{k=n}^\infty\frac{1}{k^2}\right)-\frac{1}{n^2}=\sum_{k=n+1}^\infty\frac{1}{k^2}\quad \text{ and, if $\epsilon_n=0$,}$$
$$x-x_{n}=x-x_{n-1}\leq\frac{1}{n^2}\leq\frac{1}{n+1}=\sum_{k=n+1}^\infty\frac{1}{k(k+1)}\leq \sum_{k=n+1}^\infty\frac{1}{k^2}$$
Hence, since $0\leq x-x_n\leq\sum_{k\geq n+1}\frac{1}{k^2}\to0$, we conlcude by the squeeze theorem that $\lim_n x_n=x\in Z_2$. Now, if $x\in(\zeta(2)-1,1)\cap Z_2$, then $x=\sum_{n=2}^\infty\frac{\epsilon_n}{n^2}$ since it must satisfy $\lfloor x\rfloor=0$. But this implies that $x\leq\sum_{n=2}^\infty\frac{1}{n^2}=\zeta(2)-1$, a contradiction. $\quad\Box$

Update: As pointed out by @Conrad, we can get a closed expression for $Z_p$ by defining a sequence of real numbers $(\alpha_j)_{j\geq0}$. For $j=0$, we define $\alpha_0=1$ and, for $j\geq1$, we define $\alpha_j>1$ as the unique real number $>1$ satisfying that $\sum_{n=j+1}^\infty\frac{1}{n^{\alpha_j}}=\frac{1}{j^{\alpha_j}}$ which is a root of $\zeta(x,j+1)-j^{-x}$ where $\zeta(s,a)$ is the Hurwitz zeta function. It can be proven as well that this sequence is monotonically increasing. Then, the closed expression is
$$\bbox[#DDFFCC,2pt,border: 2px solid green]{\text{if }p\in(\alpha_j,\alpha_{j+1}]\text{ then } Z_p=\left\{x+\sum_{n=1}^j\frac{\epsilon_n}{n^p}\ \bigg\vert\ \epsilon_n\in\{0,1\}\text{ and }x\in\left[0,\zeta(p,j+1)\right] \right\}}$$
where we naturally consider $\sum_{n=1}^0=0$ implying that, for $p\in(\alpha_0,\alpha_1]$, we have $Z_p=[0,\zeta(p)]$. This leads to the following formulas where $p\in(\alpha_1,\alpha_2]$ (such as $p=2$) and $q\in(\alpha_2,\alpha_3]$ (such as $q=3$)
$$Z_p=[0,\zeta(p)-1]\cup[1,\zeta(p)]$$
$$Z_q=\left[0,\zeta(q)-1-\frac{1}{2^q}\right]\cup\left[\frac{1}{2^q},\zeta(q)-1\right]\cup\left[1,\zeta(q)-\frac{1}{2^q}\right]\cup\left[1+\frac{1}{2^q},\zeta(q)\right]$$
 A: The answer to the question you asked is no as long as say $\zeta(p)-1-2^{-p}< 2^{-p}$ - for example, this happens for $p \ge 3$ by checking with Wolfram Alpha, for the simple reason that then you won't be able to get numbers close to but smaller than $1+2^{-p}$, since you will have to take $\epsilon_2=0$ and then even if $\epsilon_n=1, n \ne 2$ the most you get is $1+\sum_{n \ge 3}n^{-p}< 1+2^{-p}$ by our assumption.
There is a unique $2<p_0<3$ where $\zeta(p_0)-1-2^{-p_0}= 2^{-p_0}$, so you will get all numbers as above for $1<p \le p_0$ pretty much by the same proof (and noting that for $p$ st $\zeta(p) \ge 2$ you get the full interval $[0, \zeta(p)]$), but from $p_0$ on you will miss the interval $(\zeta(p)-2^{-p}, 1+2^{-p})$
Then if you increase $p$ to $p_1$ where $\zeta(p_1)-1-2^{-p_1}-3^{-p_1}= 3^{-p_1}$ and go beyond $p_1$ you will also miss the interval $(\zeta(p)-2^{-p}-3^{-p}, 1+3^{-p})$ for precisely the same reason, while similarly, you will miss $(\zeta(p)-3^{-p}, 1+2^{-p}+3^{-p})$
And from there it should be clear that as you increase $p$ you will miss a bunch of intervals as above for each $k$ for which $k^{-p} >\sum_{m \ge k+1}m^{-p}$
