What is known about sums of the form $\sum_{n=2}^{\infty}[\zeta(n)-1]^{p} $?

Question

A fair bit is known about rational zeta series. This includes identities like $$ \sum_{n=2}^{\infty} [\zeta(n) -1] = 1 . $$ Many more identities can be found in articles by e.g. Borwein and Adamchik & Srivastava (here).

So far, I have not been able to find identities for series involving powers of zeta values. For instance, I wonder what the collection of series $$ R(p) := \sum_{n=2}^{\infty}[\zeta(n)-1]^{p} $$ amounts to, for some positive integer $p$.

For $p=2$, we can use the first identity to establish:

\begin{align} \sum_{n=2}^{\infty} [\zeta(n)-1]^{2} &= \sum_{n=2}^{\infty} [\zeta(n)^{2} - 2 \zeta(n) + 1] \\ &= \sum_{n=2}^{\infty} (\zeta(n)^{2} - 1) -2 \sum_{n=2}^{\infty} (\zeta(n)-1) \\ &= \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) -2 .\end{align}

In order to proceed with the sum on the left, we can plug in the definition of the Riemann zeta function, use the multinomial theorem and interchange the order of summation to obtain:

\begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) &= \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} \\ \end{align}

Here, $H_{m}$ is the $m$'th Harmonic number.

Let $$S := \sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} . $$

I've considered using the following generalization of the Harmonic numbers for real and complex values $x$: $$H_{x} = \sum_{k=1}^{\infty} \binom{x}{k} \frac{(-1)^{k}}{k} $$ at $x=-\frac{1}{m}$, but I'm somewhat stuck at finding a useful expression for $\binom{-\frac{1}{m}}{k} $.

Questions:

1. Can the sum $S$ be evaluated?
2. What is known about the series $R(p)$ when $p \in \mathbb{Z}_{\geq 2}$?
3. Are there any results regarding rational sums of powers of zeta values in the literature?

Note: A copy of this question with fewer details can be found here

Answer 1

An integral representation:$$\sum (\zeta(n)^2-1)=\sum_{n\geqslant 2, (a,b)\ne (1,1)}\frac1{a^nb^n}=\sum_{(a,b)\ne (1,1)}\frac1{ab(ab-1)}\\= \sum_{(a,b)\ne (1,1)}\int_0^1 x^{ab-2}(1-x)dx=\int_0^1 (1-x)\left(\sum_{b\geqslant 2}x^{b-2}+x^{-2}\sum_{a=2}^\infty \sum_{b=1}^\infty x^{ab}\right)\\=1+ \int_0^1 \sum_{a=2}^\infty(1-x)\frac{x^{a-2}}{1-x^a}dx= 1+ \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx.$$