Timeline for Do generalizations of the identity $\sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) = 1 $ exist?

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Sep 22, 2020 at 1:41	comment	added	Alapan Das		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 $\zeta(s)$.
Sep 22, 2020 at 1:17	comment	added	Alapan Das		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.
Sep 21, 2020 at 12:29	comment	added	Max Lonysa Muller		@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?
Sep 21, 2020 at 5:14	comment	added	Alapan Das		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 $$\zeta(s)=\sum_{n=1}^{\infty} \binom{s+n-1}{n}(\zeta(s+n)-1)$$.
Sep 20, 2020 at 9:31	history	edited	Max Lonysa Muller	CC BY-SA 4.0	specified the question a bit more
Sep 19, 2020 at 18:55	history	asked	Max Lonysa Muller	CC BY-SA 4.0