Riemann zeta function related to the complete exponential bell polynomial in terms of the generalized harmonic number. Prove that
$$\zeta(s)=\frac 1 {s!(1-2^{1-s})} \sum_{n=1}^\infty \frac{B_s(0!H_n^{(1)},1!H_n^{(2)},\dots,(s-1)!H_n^{(s)})}{2^{n+1}}$$
where $B_n(x_1,\dots,x_n)$ or denoted $Y_n(x_1,\dots,x_n)$ is the $n$th complete exponential Bell polynomial and $H_n^{(k)}$ is the $n$th generalized harmonic number. 
This may likely be in the literature somewhere so a reference would suffice if found.
This converges pretty fast. Along with a proof I'd be interested to know how fast it converges.
Edit: I found the reference to the equation. See link in comment. 
 A: Start with the sum
$$\sum_{n=1}^\infty \frac{1}{2^n} 
B_q(0! H_n^{(1)}, 1! H_n^{(2)}, \ldots, (q-1)! H_n^{(q)}).$$
We have by definition that
$$B_q(x_1, x_2,\ldots x_q) = q! [w^q] 
\exp\left(\sum_{l\ge 1} x_l \frac{w^l}{l!}\right).$$
It follows that
$$B_q(0!x_1, 1!x_2,\ldots (q-1)! x_q) = q! [w^q] 
\exp\left(\sum_{l\ge 1} x_l \frac{w^l}{l}\right).$$
This is  the exponential formula (OGF  of the cycle  index $Z(S_q)$ of
the  symmetric group  using the  variables  $x_1$ to  $x_q$). We  thus
obtain for the target sum
$$q! \sum_{n=1}^\infty \frac{1}{2^n} 
\left.Z(S_q)\left(\sum_{k=1}^n \frac{1}{k^s}\right)\right|_{s=1.}$$
The  symmetric group  corresponds to  the unlabeled  multiset operator
$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{MSET}$ and hence we have from first principles
$$Z(S_q)\left(\sum_{k=1}^n \frac{1}{k^s}\right)
= [z^q] \prod_{k=1}^n 
\left(1+\frac{z}{k^s}+\frac{z^2}{k^{2s}}+\cdots\right)
= [z^q] \prod_{k=1}^n \frac{1}{1-z/k^s}.$$
We  extract the coefficient  using partial  fractions by  residues and
write (setting $s=1$)
$$[z^q] \prod_{k=1}^n \frac{k}{k-z}
= (-1)^n [z^q] \prod_{k=1}^n \frac{k}{z-k}
= (-1)^n n! [z^q] \prod_{k=1}^n \frac{1}{z-k}.$$
We get for the residue at $z=p$
$$(-1)^n n! \prod_{k=1}^{p-1} \frac{1}{p-k}
\prod_{k=p+1}^{n} \frac{1}{p-k}
= (-1)^n n! \frac{1}{(p-1)!} \frac{(-1)^{n-p}}{(n-p)!}
\\ = (-1)^p p {n\choose p}.$$
Extracting coefficients now yields
$$[z^q] \sum_{p=1}^n (-1)^p p {n\choose p} \frac{1}{z-p}
= - [z^q] \sum_{p=1}^n (-1)^p {n\choose p} \frac{1}{1-z/p}
\\ = - \sum_{p=1}^n (-1)^p {n\choose p} \frac{1}{p^q}.$$
Substitute into the sum to get
$$- q! \sum_{n=1}^\infty \frac{1}{2^n}
\sum_{p=1}^n (-1)^p {n\choose p} \frac{1}{p^q}
= - q! \sum_{p\ge 1} (-1)^p \frac{1}{p^q}
\sum_{n\ge p} {n\choose p} \frac{1}{2^n}
\\ = - q! \sum_{p\ge 1} (-1)^p \frac{1}{p^q} \frac{1}{2^p}
\sum_{n\ge 0} {n+p\choose p} \frac{1}{2^n}
\\ = - q! \sum_{p\ge 1} (-1)^p \frac{1}{p^q} \frac{1}{2^p}
\frac{1}{(1-1/2)^{p+1}}
\\ = - 2q! \sum_{p\ge 1} (-1)^p \frac{1}{p^q} 
= 2q! \left(1-\frac{1}{2^q} + \frac{1}{3^q} - \cdots\right)
= 2q! \times \zeta(q) \times \left(1-\frac{2}{2^q}\right)
\\ = 2q! \times \zeta(q) \times (1-2^{1-q}).$$
This is the claim and we are done. This computation is closely related
to     what     was     presented     at    the     following     MSE
link.
