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

Can the identity be generalised and/or are there variations of it involving binomial sums which are also valid for infinitely many nonnegative lower bounds $k$?

I suppose one can rewrite the formulas of Landau: $$ \frac{1}{s-1} = \sum_{n=0}^{\infty} \binom{s+n-1}{n-1} \frac{\zeta(s+n)-1}{n} $$ and Ramaswami: $$ (1-2^{1-s})\zeta(s) = \sum_{n=1}^{\infty} \binom{s+n-1}{n} \zeta(s+n)$$ in a manner that is similar to $(*)$, but I'm looking for both more and more general examples of such sums. Preferably, all sums in the collection amount to the same constant and involve binomial coefficients.

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  • $\begingroup$ These later identities are wrong. The first sum should be be $$\frac{\zeta(s)}{s}=\sum_{n=0}^{\infty} \binom{n+s-1}{n-1} \frac{(\zeta(s+n)-1)}{n}$$. And the second identity isn't converging. That may be $$\zeta(s)=\sum_{n=1}^{\infty} \binom{s+n-1}{n}(\zeta(s+n)-1)$$. $\endgroup$
    – Alapan Das
    Commented Sep 21, 2020 at 5:14
  • $\begingroup$ @AlapanDas Hmm, these identities were retrieved from p. 286 of Borwein's paper. So if they're wrong, you've found an error in a peer-reviewed article. Do you have sources for your expressions? $\endgroup$
    – Max Lonysa Muller
    Commented Sep 21, 2020 at 12:29
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    $\begingroup$ These identities can be proved by the identity $\zeta(n)=\frac{1}{\Gamma(n)}\int_{0}^{\infty} \frac{x^{n-1}}{e^x-1} dx$. Also, I have checked these identities in Wolfram alpha. $\endgroup$
    – Alapan Das
    Commented Sep 22, 2020 at 1:17
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    $\begingroup$ In my first comment, I have made a small error in the second identity. If we take $n$ from $1$, then the value is $1$, if the limit is from $n=0$, then the value will be $\zeta(s)$. $\endgroup$
    – Alapan Das
    Commented Sep 22, 2020 at 1:41

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