3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.