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A fair bit is known about rational zeta series. This includes identities like $$ \sum_{n=2}^{\infty} [\zeta(n) -1] = 1 . $$ Many more identities can be found in articles by e.g. Borwein and Adamchik & Srivastava (here).

So far, I have not been able to find identities for series involving powers of zeta values. For instance, I wonder what the collection of series $$ R(p) := \sum_{n=2}^{\infty}[\zeta(n)-1]^{p} $$ amounts to, for some positive integer $p$.

For $p=2$, we can use the first identity to establish:

\begin{align} \sum_{n=2}^{\infty} [\zeta(n)-1]^{2} &= \sum_{n=2}^{\infty} [\zeta(n)^{2} - \zeta(n) + 1] \\ &= \sum_{n=2}^{\infty} (\zeta(n)^{2} - 1) -2 \sum_{n=2}^{\infty} (\zeta(n)-1) \\ &= \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) -2 .\end{align}

In order to proceed with the sum on the left, we can plug in the definition of the Riemann zeta function, use the multinomial theorem and interchange the order of summation to obtain:

\begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) &= \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} \\ \end{align}

Here, $H_{m}$ is the $m$'th Harmonic number.

Let $$S := \sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} . $$

I've considered using the following generalization of the Harmonic numbers for real and complex values $x$: $$H_{x} = \sum_{k=1}^{\infty} \binom{x}{k} \frac{(-1)^{k}}{k} $$ at $x=-\frac{1}{m}$, but I'm somewhat stuck at finding a useful expression for $\binom{-\frac{1}{m}}{k} $.

Questions:

  1. Can the sum $S$ be evaluated?
  2. What is known about the series $R(p)$ when $p \in \mathbb{Z}_{\geq 2}$?
  3. Are there any results regarding rational sums of powers of zeta values in the literature?

Note: A copy of this question with fewer details can be found here

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  • $\begingroup$ I recomond you if you ask again you should linked your related questions $\endgroup$ – zeraoulia rafik Oct 3 '20 at 21:31
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    $\begingroup$ @zeraouliarafik Will do, I forgot this time. $\endgroup$ – Max Muller Oct 3 '20 at 21:31
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    $\begingroup$ for what it worth, $\sum (\zeta(n)^2-1)=1+\int_0^1 \sum_{k=0}^\infty \frac{x^k}{1+x+x^2+\ldots+x^{k+1}} dx$ $\endgroup$ – Fedor Petrov Oct 3 '20 at 23:39
  • $\begingroup$ @FedorPetrov thank you for this insight. To me, it's not immediately obvious that the equality holds. Could you please elaborate on it? $\endgroup$ – Max Muller Oct 4 '20 at 12:45
  • $\begingroup$ @MaxMuller did it in the answer $\endgroup$ – Fedor Petrov Oct 4 '20 at 15:25
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An integral representation:$$\sum (\zeta(n)^2-1)=\sum_{n\geqslant 2, (a,b)\ne (1,1)}\frac1{a^nb^n}=\sum_{(a,b)\ne (1,1)}\frac1{ab(ab-1)}\\= \sum_{(a,b)\ne (1,1)}\int_0^1 x^{ab-2}(1-x)dx=\int_0^1 (1-x)\left(\sum_{b\geqslant 2}x^{b-2}+x^{-2}\sum_{a=2}^\infty \sum_{b=1}^\infty x^{ab}\right)\\=1+ \int_0^1 \sum_{a=2}^\infty(1-x)\frac{x^{a-2}}{1-x^a}dx= 1+ \int_0^1 \sum_{k=0}^\infty\frac{x^{k}}{1+x+\ldots+x^{k+1}}dx.$$

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    $\begingroup$ Thank you! That is a beautiful representation, in my opinion. Also, if we interchange the sum and the integral sign and evaluate the integral, of each term, it is equal to $ - \frac{H_{-\frac{1}{k+2}}}{k+2}$, so now there's only one sum to evaluate. Do you have any ideas on or references to sums of negative fractional harmonic numbers? $\endgroup$ – Max Muller Oct 4 '20 at 17:08
  • $\begingroup$ Sleek ! Very sleek ! $\endgroup$ – Aditya Guha Roy Oct 6 '20 at 17:16

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