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.

  • $\begingroup$ you mean $\Gamma(s)$ instead of $s!$ ? Then $(1-2^{1-s}) \zeta(s)\Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^x+1}dx$ $\endgroup$ – reuns May 5 '17 at 5:37
  • $\begingroup$ I don't know what is supposed to be $B_s$ for $s \not \in \mathbb{N}$. If you meant $s \in \mathbb{N}$ then use the definition of $B_n$ and $H_n^{(s)}$ $\endgroup$ – reuns May 5 '17 at 5:43
  • $\begingroup$ Yes, we can let $s\in\mathbb{N}$ $\endgroup$ – tyobrien May 5 '17 at 12:51
  • $\begingroup$ I found the reference. arxiv.org/pdf/1001.2835.pdf equation 3.23. But my question on rate of convergence still stands. $\endgroup$ – tyobrien May 6 '17 at 0:49

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 $\mathfrak{M}$ 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.

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  • $\begingroup$ Instructive solution! Good source to improve my techniques (+1) $\endgroup$ – Markus Scheuer May 14 '17 at 6:16

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