As per <a href="https://mathoverflow.net/questions/373231/what-is-known-about-sums-of-the-form-sum-n-2-infty-zetan-1p">these</a> <a href="https://mathoverflow.net/questions/373499/a-question-on-an-identity-relating-certain-sums-of-harmonic-numbers">questions</a>, I'm trying to evaluate $$\sum_{n=2}^{\infty} \big{(} \zeta(n)^{2}-1 \big{)} = 1+ \sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m}. $$
Here, $H_{x}$ is a generalized <a href="https://en.wikipedia.org/wiki/Harmonic_number#Generalizations">Harmonic</a> number. When we proceed with the definition $$\begin{align} H_{x} = \sum_{k=1}^{\infty} \binom{x}{k} \frac{(-1)^{k}}{k} \end{align} $$ and consider the <a href="https://math.stackexchange.com/questions/3851180/references-on-binomial-coefficients-of-the-form-binom-1-mk">fact</a> that relates this expression to the *unsigned Stirling numbers of the first kind*: $$\begin{align} (-1)^n {-m \choose n} = {n+m-1 \choose n} = \frac{1}{n!} \sum_{i=0}^n \left[ {n \atop i} \right] m^i , \end{align} $$ we can replace $m$ by $\frac{1}{m}$ and rearrange sums to obtain:
$$\begin{align}
\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} &= \sum_{m=2}^{\infty} \sum_{k=1}^{\infty} \sum_{i=0}^{k} \frac{1}{k k!} m^{-(i+1)} \left[ {k \atop i} \right] \\
&= \sum_{k=1}^{\infty} \sum_{i=0}^{k} \frac{1}{k k!} \left[ {k \atop i} \right] \sum_{m=2}^{\infty} m^{-(i+1)} \\
&= \sum_{k=1}^{\infty} \sum_{i=0}^{k} \frac{1}{k k!} \left[ {k \atop i} \right] (\zeta(i+1) - 1) \\
&= \bigg{(} \sum_{k=1}^{\infty} \frac{1}{kk!} \bigg{(} \sum_{i=0}^{k} \left[ {k \atop i} \right] \zeta(i+1) - \sum_{i=0}^{k} \left[ {k \atop i} \right] \bigg{)} \bigg{)} \\
&= \bigg{(} \sum_{k=1}^{\infty} \frac{1}{kk!} \bigg{(} \sum_{i=1}^{k} \left[ {k \atop i} \right] \zeta(i+1) - k! \bigg{)} \bigg{)}.
\end{align}.$$
In the last equation, I assumed that we could write $\left[ {k \atop 0} \right] \zeta(1)=0 $ because $\left[ {k \atop 0} \right] = 0$. If this isn't true, we may have to <a href="https://math.stackexchange.com/questions/4316335/what-is-lim-t-to-0-left-k-atop-t-right-zetat1">compute</a> $\lim_{t \to 0} \left[ {k \atop t} \right] \zeta(t+1) $ by means of some notion of the analytic <a href="https://www.tandfonline.com/doi/abs/10.1080/10236190902780190">continuation</a> of the Stirling numbers.
The focus of this question isn't so much on this boundary case, however, but on expressions for $$\sum_{i=1}^{k} \left[ {k \atop i} \right] \zeta(i+1). \tag{*} $$
I've tried finding instances of this sum in the literature, but so far I only found information on sums that are related to the sum $$\sum_{i=1}^{n}{n \brack i}\zeta(n-i) \tag{**}$$
For instance, I found:
1. A relation between $(**)$ and the hypergeometric function on p. 123 of a <a href="https://www.sciencedirect.com/science/article/pii/S0377042796001677">paper</a> by Adamchik: $$ \begin{align}
{}_{p+1}F_{p} {k, k, \dots, k \choose k+1, \dots, k+1; 1} = \frac{k^{p}}{(k-1)!} \sum_{i=0}^{k-1}(-1)^{k-i-1} \zeta(p-i) \left[ {k \atop i+1} \right].
\end{align} \tag{1} $$
2. If we <a href="https://mathoverflow.net/questions/262053/multiple-integral-evaluation-involving-stirling-numbers-and-riemann-zeta-functio">define</a> $$K_n(m) = \overbrace{\int_0^1 \dots \int_0^1}^{n-\mathrm{times}} \left(-\frac{\ln(1-x_1x_2\cdots x_n)}{x_1 x_2 \cdots x_n}\right)^m \mathrm{d}x_1\mathrm{d}x_2 \cdots \mathrm{d}x_n , $$ then $$K_1(m) = m\sum_{n=0}^{m-1}\left[ m-1 \atop n\right]\zeta(m+1-n) . \tag{2}$$
3. For all integer $n>m\ge0$, we <a href="https://math.stackexchange.com/questions/2733728/closed-form-for-int-01-frac-lognx1-xm-dx">have</a>:
$$
S(n,m):=\int_0^1\frac{\log^n(1-u)}{u^{m+1}}du=\frac{(-1)^n n!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i). \tag{3}
$$
However, I haven't found a single expression yet for $(*)$.
I thus have two **questions**:
> 1. Is there a particular reason why expressions related to the form in $(**)$ seem more prevalent in mathematical research than identities
> pertaining to $(*)$ ?
> 2. Even more importantly to me: is there any literature on sums of the form $(*)$ ?