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As per these questions, 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 Harmonic 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 fact 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 compute $\lim_{t \to 0} \left[ {k \atop t} \right] \zeta(t+1) $ by means of some notion of the analytic continuation 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 paper 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 define $$K_n(m) = \overbrace{\int_0^1 \dots \int_0^1}^{n-\mathrm{times}} \left(-\frac{\ln(1-x_1x_2\cdots x_n)}{x_1 x_2 \cdots x_n}\right)^m \mathrm{d}x_1\mathrm{d}x_2 \cdots \mathrm{d}x_n , $$ 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 have: $$ S(n,m):=\int_0^1\frac{\log^n(1-u)}{u^{m+1}}du=\frac{(-1)^n n!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i). \tag{3} $$

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 $(*)$ ?
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  • $\begingroup$ Should your first formula really be $\sum\limits_{k=2}^\infty\left(\zeta(k)^{2}-1 \right)=1+\sum\limits_{m=2}^\infty\frac{H_{-\frac{1}{m}}}{m}$? (changed $\zeta(n)$ to $\zeta(k)$ on left-side, added $1$ and changed sign on right-side). Also I believe $H_x$ has the closed form $H_x=-\psi ^{(0)}(x+1)-\gamma$. $\endgroup$ Nov 28, 2021 at 17:15
  • $\begingroup$ @StevenClark Yes, you're right about the first formula - I'll change it quickly, thank you. The harmonic numbers are indeed related to the digamma function as you described, but I'm afraid this relationship doesn't help much with the evaluation of the sum $\endgroup$
    – Max Muller
    Nov 28, 2021 at 17:21

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