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It is well known and easy to evaluate, that $\sum_{n=2}^{\infty}\left(\zeta (n)-1\right)=1$. I am trying to evaluate $$\sum_{n=2}^{\infty}\left(\frac{1}{\zeta (n)}-1\right)$$ The classical way gives me $$\sum_{n=2}^{\infty}\left(\frac{1}{\zeta (n)}-1\right)=\sum_{n=2}^{\infty}\frac{\mu(n)}{n\cdot(n-1)}=-1+\sum_{n=2}^{\infty}\frac{\mu(n+1)-\mu(n)}{n}=1+\sum_{n=2}^{\infty}\frac{\mu(n)}{n-1}$$ where $\mu$ is the Mobius function. But it dosn't help me at all.

WolframAlpha gives the result -0.705211 (exactly). EDIT The value -0.705211 is not exact, although WolframAlpha gives = and not ≈. A more accurate value is -0.705211140...

Q: How to evaluate $\sum_{n=2}^{\infty}\left(\frac{1}{\zeta (n)}-1\right)$?

Thank you very much.

zhrd
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    "WolframAlpha gives the result -0.705211 (exactly)." - why did you decide it's the exact value? It's not – Yuriy S Dec 06 '23 at 09:06
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    Because the answer is sum_(n=2)^∞(1/ζ(n) - 1) = -0.705211 and not ≈-0.705211 – zhrd Dec 06 '23 at 09:15
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    You have no idea how Wolfram Alpha works, do you? If you refuse to be corrected, I doubt anyone can help you – Yuriy S Dec 06 '23 at 09:17
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    My question was how to evaluate the sum and not how Wolfram Alpha works. – zhrd Dec 06 '23 at 09:19
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    Because what you claim to be the exact value of the sum is not its exact value. I don't know how else to explain it to you – Yuriy S Dec 06 '23 at 09:33
  • What makes you think that this sum has a nice value, namely nicer than $\sum_{n=2}^{+\infty}\left(\frac{1}{\zeta(n)}-1\right)$ ? – Tuvasbien Dec 06 '23 at 16:20
  • To be honest, nothing. In the meantime, I found the Niven's Constant, and probably a "nicer" value even doesn't exist. Sorry for the stupid question. – zhrd Dec 07 '23 at 04:10
  • @zhrd That's not a stupid question, Niven himself wrote a paper about this, and he's a professional mathematician ! – Tuvasbien Dec 08 '23 at 00:38

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