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Oct 30, 2021 at 20:52 history edited Max Lonysa Muller CC BY-SA 4.0
Corrected a small mistake in a formula
May 26, 2021 at 12:39 comment added Max Lonysa Muller @StevenClark maybe! It could work, I'm not sure
May 23, 2021 at 14:42 comment added Steven Clark Would the representation $H_s=\sum\limits_{n=1}^\infty\frac{s}{n\ (s+n)}$ be useful? More generally $H_s^{(r)}=\sum\limits_{n=1}^\infty\left(\frac{1}{n^r}-\frac{1}{(n+s)^r}\right)=\zeta (r)-\zeta (r,s+1)$.
Oct 9, 2020 at 12:14 history edited Max Lonysa Muller CC BY-SA 4.0
corrected an error in a calculation
Oct 4, 2020 at 15:25 comment added Fedor Petrov @MaxMuller did it in the answer
Oct 4, 2020 at 15:24 answer added Fedor Petrov timeline score: 7
Oct 4, 2020 at 15:01 history edited Max Lonysa Muller CC BY-SA 4.0
corrected grammar
Oct 4, 2020 at 12:45 comment added Max Lonysa Muller @FedorPetrov thank you for this insight. To me, it's not immediately obvious that the equality holds. Could you please elaborate on it?
Oct 3, 2020 at 23:39 comment added Fedor Petrov for what it worth, $\sum (\zeta(n)^2-1)=1+\int_0^1 \sum_{k=0}^\infty \frac{x^k}{1+x+x^2+\ldots+x^{k+1}} dx$
Oct 3, 2020 at 21:46 history edited Max Lonysa Muller CC BY-SA 4.0
edited body
Oct 3, 2020 at 21:31 comment added Max Lonysa Muller @zeraouliarafik Will do, I forgot this time.
Oct 3, 2020 at 21:31 comment added zeraoulia rafik I recomond you if you ask again you should linked your related questions
Oct 3, 2020 at 21:26 history edited zeraoulia rafik CC BY-SA 4.0
added 106 characters in body
Oct 3, 2020 at 18:55 history asked Max Lonysa Muller CC BY-SA 4.0