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I wonder whether $\gamma$ belongs to the ring of periods?

UPDATE Well now I know it should not. But $e^{-\gamma}$ should.

Anixx
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  • The irrationality of $\gamma$ is a long-standing conjecture. However, it is widely believed that $\gamma\not\in\mathbb{Q}$. – Jack D'Aurizio Jan 04 '15 at 12:32
  • This is an open question. – Lucian Jan 04 '15 at 13:06
  • Does the existence of the following series help to solve the question? $$ \gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right) $$ http://math.stackexchange.com/a/1591256/134791 – Jaume Oliver Lafont Jan 20 '16 at 06:27

1 Answers1

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On the one hand, it is expected not to be a period. On the other hand, it could even be rational. So, we do not know for sure that it is not a period.

quid
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  • Expected by whom not to be period? – Anixx Jan 04 '15 at 12:18
  • The issue is, AFAIK, that $e$ is expected not to be a period either. Even the idea of adding $e$ was entertained to define "exponential periods" and there you might have it. – quid Jan 04 '15 at 12:31
  • Also it says on Wikipedia "It is not expected that Euler's number e and Euler–Mascheroni constant γ are periods. The periods can be extended to exponential periods by permitting the product of an algebraic function and the exponential function of an algebraic function as an integrand. This extension includes all algebraic powers of e, the gamma function of rational arguments, and values of Bessel functions. If, further, Euler's constant is added as a new period, then according to Kontsevich and Zagier "all classical constants are periods in the appropriate sense"." – quid Jan 04 '15 at 12:31
  • Regarding $e$, it is clear not a period, there is little doubt. Not the case with $\gamma$ though. Even more than $\gamma$, $e^{-\gamma}$ is likely to be a period. – Anixx Jan 04 '15 at 12:43
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    @Anixx, in what possible sense is it «clear» that e is not a period?!?!?! – Mariano Suárez-Álvarez Jan 20 '16 at 08:07