Some questions on the identity $\sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) = 1 $

Question

On p. 263 of Borwein's paper entitled “Computational Strategies for the Riemann zeta function”, the following identity is stated: $$\sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) =1 .$$It is valid for all nonnegative integer $k$.

The author points to other formulas that are relevant, including (43) on p. 262. However, I don't see how this identity follows from any previously established series evaluations, nor have I found a reference in the article with regards to this particular identity.

Questions

1. How can the aforementioned identity be derived? Can you provide a proof?
2. Are there any references to articles or books that describe and prove this identity?
3. Can the identity be generalised? Are there variations of it involving binomial sums which are also valid for all nonnegative $k$?

Answer 1

$$ \sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) = \sum_{n=k+2}\binom{n-1}{k}\sum_{m=2}^{\infty} \frac{1}{m^{n}} \\ = \sum_{m=2}^{\infty} \sum_{n=k+2}^{\infty} \binom{n-1}{k}\frac{1}{m^{n}} $$ From $$\binom{n-1}{k} = (-1)^{n-1-k}\binom{-1-k}{n-1-k}$$ we have $$ (sum) = \sum_{m=2}^{\infty}\sum_{n=k+2}^{\infty} (-1)^{n-1-k}\binom{-1-k}{n-1-k}\frac{1}{m^{n}} \\ =\sum_{m=2}^{\infty}\frac{1}{m^{k+1}}\sum_{n=1}^{\infty}\binom{-k-1}{n}\left(-\frac{1}{m}\right)^{n} \\ =\sum_{m=2}^{\infty}\frac{1}{m^{k+1}}\left[\left(1 - \frac{1}{m}\right)^{-k-1} - 1\right]\qquad(\text{by Newton's binomial theorem}) \\ =\sum_{m=2}^{\infty}\left(\frac{1}{(m-1)^{k+1}} - \frac{1}{m^{k+1}}\right) = 1. $$

Answer 2

1. Using the integral representation of the Riemann Zeta function we have that $$\sum_{n=2}^{\infty}\dbinom{n+k-1}{k}\left(\zeta\left(n+k\right)-1\right)=\int_{0}^{+\infty}\sum_{n=2}^{\infty}\frac{u^{n+k-1}}{k!\left(n-1\right)!}\left(\frac{1}{e^{u}-1}-\frac{1}{e^{u}}\right)du$$ $$=\frac{1}{k!}\int_{0}^{+\infty}u^{k}e^{-u}du=1.$$

2. I don't know references to this identity.

3. Note that, in this formulation, the results may be generalized for all $k\in\mathbb{R}^{+}$, obviously taking $k!=\Gamma\left(k+1\right)$. About the “variations” I'm not sure I understand what you are asking: do you mean sums with Zeta function and binomials? Or parametric sums with Zeta and binomials such that the result is constant? However, in the link I posted there are a lot of sum involving Zeta function.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\sum_{n\ =\ k + 2}^{\infty} {n - 1 \choose k}\bracks{\zeta\pars{n} - 1} = 1}:\ {\Large ?}}$ Note that \begin{align} &\sum_{n = k + 2}^{\infty}{n - 1 \choose k} = \bracks{z^{k}}\sum_{n = k + 2}^{\infty} \pars{1 + z}^{n - 1} = \bracks{z^{k}}{\pars{1 + z}^{k + 1} \over 1-\pars{1 + z}} \\[5mm] = &\ -\bracks{z^{k + 1}}\pars{1 + z}^{k + 1} = \bbx{\large -1} \\ & \end{align} \begin{align} &\mbox{such that} \bbox[5px,#ffd]{\sum_{n\ =\ k + 2}^{\infty} {n - 1 \choose k}\bracks{\zeta\pars{n} - 1}} \\[5mm] = &\ 1\ +\ \underbrace{\sum_{n\ =\ k + 2}^{\infty} {n - 1 \choose k}\zeta\pars{n}} _{\ds{=\ \color{red}{\large 0}}}\ =\ \bbx{\Large 1} \\ & \end{align} $\ds{\bbx{\mbox{It remains to prove that}\ \sum_{n\ =\ k + 2}^{\infty} {n - 1 \choose k}\zeta\pars{n} = \color{red}{\large 0}}}$.

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With A & S $\ds{\bf\color{black}{6.3.14}}$ identity $\ds{\pars{~\left.\vphantom{\large A}\Psi\pars{z} \right\vert_{\ \verts{z}\ <\ 1} = -\gamma + \sum_{n = 2}^{\infty}\pars{-1}^{n}\zeta\pars{n}z^{n - 1}~}}$: \begin{align} &\bbox[5px,#ffd]{\sum_{n = k + 2}^{\infty} {n - 1 \choose k}\zeta\pars{n}} \\[5mm] = &\ \sum_{n = k + 2}^{\infty} {n - 1 \choose k} \bracks{z^{n - 1}}\braces{\pars{-1}^{n} \bracks{\Psi\pars{z} + \gamma}} \\[5mm] = &\ -\bracks{z^{0}}\bracks{\Psi\pars{z} + \gamma} \sum_{n = k + 2}^{\infty} \pars{-\,{1 \over z}}^{n - 1}{n - 1 \choose k} \\[5mm] = &\ -\bracks{z^{0}}\bracks{\Psi\pars{z} + \gamma} \sum_{n = 1}^{\infty} \pars{-\,{1 \over z}}^{n + k}{n + k \choose k} \\[5mm] = &\ \pars{-1}^{k + 1} \bracks{z^{k}}\bracks{\Psi\pars{z} + \gamma} \sum_{n = 1}^{\infty} \pars{-\,{1 \over z}}^{n}{n + k \choose n} \\[5mm] = &\ \pars{-1}^{k + 1} \bracks{z^{k}}\bracks{\Psi\pars{z} + \gamma} \sum_{n = 1}^{\infty} \pars{-\,{1 \over z}}^{n} {- k - 1 \choose n}\pars{-1}^{n} \\[5mm] = &\ \pars{-1}^{k + 1} \bracks{z^{k}}\bracks{\Psi\pars{z} + \gamma} \bracks{\pars{1 + {1 \over z}}^{-k - 1} - 1} \\[5mm] = &\ \pars{-1}^{k + 1} \bracks{z^{k}}\bracks{\Psi\pars{z} + \gamma} \bracks{{z^{k + 1} \over \pars{1 + z}^{k + 1}} - 1} \\[5mm] = &\ \bbx{\large 0} \\ & \end{align}