On p. 286 of Borwein's [paper](https://cr.yp.to/bib/2000/borwein.pdf) entitled "Computational Strategies for the Riemann zeta function", the author mentions a formula due to Ramaswami: $$(1-2^{1-s})\zeta(s) = \sum\_{n=1}^{\infty} \binom{s+n-1}{n}\zeta(s+n). $$
I wonder whether variations of this identity also exist. For instance, are there similar binomial sums for $$(1-a^{1-s})\zeta(s) $$ for $a \in \mathbb{Z}\setminus\{2\}$, or is there something special about $a=2$ that makes it work?


And what about products like $$\zeta(s) \prod\_{k=1}^{p} (1-a\_{k}^{1-s})$$ for some sequence $a\_{1}, \dots, a\_{p} \in \mathbb{Z}$, does that expression equal any binomial sum(s) in terms of values of the Riemann zeta function?




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N.B. I've also asked this [question](https://math.stackexchange.com/questions/3820251/are-there-variations-of-ramaswamis-formula-for-the-analytic-continuation-of-the) on MSE.




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I've corrected some typos. The $2^{-s}$, $a^{-s}$, and $a\_{k}^{-s}$ factors should have been $2^{1-s}$, $a^{1-s}$, and $a\_{k}^{1-s}$, respectively.



Are there variations of Ramaswami's formula for the analytic continuation of the Riemann zeta function?
First note that there is a typo in the formula you cite: it should be
$$ (1-2^{1-s})\zeta(s) = \sum\_{n=1}^{\infty} \binom{s+n-1}{n}\zeta(s+n) $$
($1-s$, not $-s$). Something "special" in the number $2$ can be found, since $(1-2^{1-s})\zeta(s) = \eta(s)$ (Dirichlet eta function). However, the above formula can be generalised for a general integer $a \geq 2$. I will refer to the paper *Lee, H.; Ok, B. M.; Choi, J. Notes on some identities involving the Riemann Zeta function (2002). Communications of the Korean Mathematical Society 17(1):165-173* for such identities.


First of all, we have the following identities obtained by Ramaswami:


$$ (1-3^{1-s})\zeta(s) = 1+2 \sum\_{n=1}^{\infty} \frac{(s)\_{2n}}{(2n)!} \zeta(2n+s) 3^{-2n-s} $$


$$ (1-2^{-s}-3^{-s}-6^{-s})\zeta(s) = 1 +2 \sum\_{n=1}^{\infty} \frac{(s)\_{2n}}{(2n)!} \zeta(2n+s) 6^{-2n-s}$$


The first one is the case $a=3$ of your first question, while the second one is a special case of the analogous of your second question, but with a sum instead of a product.


The general case for any integer $a \geq 2$ has been proven by Apostol. Actually, he obtained many interesting identities of this kind:


$$ (1-a^{1-s})\zeta(s) = \sum\_{n=1}^{\infty} \frac{(s)\_{n} \zeta(n+s)}{n! a^{n+s}} \frac{B\_{n+1}(a) - B\_{n+1}}{n+1} $$


$$ (1-a^{1-s})\zeta(s) = \sum\_{h=1}^{a-1} h^{-s} + \sum\_{n=1}^{\infty} (-1)^n \frac{(s)\_{n} \zeta(n+s)}{n! a^{n+s}} \frac{B\_{n+1}(a) - B\_{n+1}}{n+1} $$


$$ (1-a^{1-s})\zeta(s) = \frac{1}{2} \sum\_{h=1}^{a-1} h^{-s} + \sum\_{n=1}^{\infty} \frac{(s)\_{2n} \zeta(2n+s)}{(2n)! a^{2n+s}} \frac{B\_{2n+1}(a)}{2n+1} $$


I'm not aware of any formula of this kind for a product as the one of your second question.


