An interesting way of expressing any real number using the harmonic series.

Some notation first: $$\mbox{HSN}(z)=H_z=\{a_0,a_1,\dots\}=0.a_0a_1\dots, \; \forall a_i. |a_i|=1$$ $$ z = \sum_{i=0}^\infty{\frac{c_i}{i+1}}, \; c_i = \begin{cases} +1 & \mbox{if } a_i=1 \\ -1 & \mbox{if } a_i=0 \end{cases} $$

Now we can write $z$ as, $$(1) \Rightarrow z =\int_0^1P_z(x)dx,\; P_z(x) = {\sum_{i=0}^\infty{c_ix^i}dx}$$

Rational HSN

define $R_{z} = \{p\in \mathbb{N}_0|\forall a_i \in H_z. a_i=a_{i+p}\}$ and $r_z = min(R_z)$ then is $H_z$ rational $\Leftrightarrow R_z \ne \varnothing$

$$ \begin{align*} H_z \text{ is rational} \Rightarrow & P_z(x)&& = {\sum_{i=0}^\infty{c_ix^i}} \\ & && = {\sum_{i=0}^\infty{c_{(i\mbox{ mod }r_z)}x^i}} \\ & && = {\sum_{i=0}^{r_z-1}{c_ix^i}} + {\sum_{i=r_z}^\infty{c_{(i\mbox{ mod }r_z)}x^i}} \\ & && = {\sum_{i=0}^{r_z-1}{c_ix^i}} + x^{r_z}{\sum_{i-r_z=0}^\infty{c_{(i-r_z\mbox{ mod }r_z)}x^{i-r_z}}} \\ & && = p_z(x) + x^{r_z}P_z(x), \; p_z(x)={\sum_{i=0}^{r_z}{c_ix^i}}\\ \Leftrightarrow& P_z(x) && = \frac{p_z(x)}{1-x^{r_z}}\\ \Leftrightarrow& z && = \int_0^1\frac{p_z(x)}{1-x^{r_z}}dx\\ \end{align*}$$

convergence

$p_z(1) \neq 0$ will cause $\frac{p_z(x)}{1-x^{r_z}}$ to converge to $\frac1x$ which has infinite area in $[1-\epsilon,1]$. So in order for $z$ to have a finite value $ p_z(1) $ must be $0$, or $\sum_{i=0}^{r_z-1}{c_i} = 0$. This also means that $r_z$ must be even.

closed form

we can solve the integral for $z$ using the partial fraction decomposition of $\frac{p_z(x)}{1-x^{r_z}}$

$$\begin{align*} (2) \Rightarrow && \frac{p_z(x)}{1-x^{r_z}} & = \sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z(x-\alpha_t)}}, \; \alpha_t = e^{2\pi ti/{r_z}}\\ \Rightarrow && z &= \int_0^1{\sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z(x-\alpha_t)}}dx}\\ && z &= \sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln\left(\frac{1-\alpha_t}{0-\alpha_t}\right)} \\ && z &= \sum_{t=1}^{r_z}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln(1-\bar\alpha_t)} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \sum_{t=1}^{r_z/2-1}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln(1-\bar\alpha_t)+\frac{p_z(\bar\alpha_t)\bar\alpha_t}{r_z}\cdot\ln(1-\alpha_t)} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \sum_{t=1}^{r_z/2-1}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\overline{\ln(1-\alpha_t)}+\frac{\overline{p_z(\alpha_t)\alpha_t}}{r_z}\cdot\ln(1-\alpha_t)} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \sum_{t=1}^{r_z/2-1}{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\overline{\ln(1-\alpha_t)}+\overline{\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\overline{\ln(1-\alpha_t)}}} \\ && z &= \frac{-p_z(-1)}{r_z}\cdot\ln(2) + \frac2{r_z}\sum_{t=1}^{r_z/2-1}{Re\left(p_z(\alpha_t)\alpha_t\cdot\overline{\ln(1-\alpha_t)}\right)} \\ \end{align*}$$ (!!) because we assumed $z$ is finite the partial fraction for $\alpha_0$ can be omitted

Question 1;

I suspect that only transcendental numbers have a rational HSN, because they can be writen as a sum of natural logaritm of algebraic numbers, so the sum of transcendental numbers. However not all sums of transcendental numbers are transcendental, you'd have to prove that ${\frac{p_z(\alpha_t)\alpha_t}{r_z}\cdot\ln(1-\bar\alpha_t)}$ are algebraically independent.

Question 2:

as far as i found not, the opposite would be easier (if Question 1)

Question 3:

If Question 1 is correct then no algebraic number could have a rational HSN

Question 4:

If Question 1 is correct, such a number would have to be irrational.

Question 5:

For rational HSN see above, for irrational HSN i suspect the same must be true.

$(1)$ $$\frac1n=\int_0^1{x^{n-1}dx}$$ $(2)$ $$\frac{P(x)}{Q(x)}=\sum_{i=0}^n{\frac{P(\alpha_i)}{Q'(\alpha_i)}\frac1{x-\alpha_i}}$$