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I got a serie for you, but if you don't like inform me and I will delete it. Let $$A=\left(1+{1\over \pi} \right)\left(6-\ln{27 }\right)-{\pi \over \sqrt3}$$ Then we have: $$\gamma+{1\over \sqrt3}=A+\sum_{n=1}^{\infty}\left({2+{2\over \pi} \over n+{1\over 3}}-{1+{2\over \pi} \over n}-\ln{n+1\over n}\right)$$


Interesting question. We may start with: $$ \gamma = \sum_{n\geq 1}\frac{1}{n}\cdot[x^n]\frac{x}{\log(1-x)}=\sum_{n\geq 1}\frac{G_n}{n}. \tag{1}$$ Due to Steffensen's bounds, we know that the Gregory coefficients $[x^n]\frac{x}{\log(1-x)}$ behave like $\frac{1}{n\log^2 n}$, so we may try to apply the Cauchy-Schwarz inequality to $(1)$ in the trivial way, ...

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