I found this formula for the Euler-Mascheroni constant $\gamma$. 



> 
> Just wondering whether such a formula already exists in literature?
>  Also, wanted to know whether there are formulas that converge faster than this?
> 
> 
> 


$$\gamma = \sum\_{k = 1}^{\infty} \frac{1}{2^k k} - \sum\_{k = 1}^{\infty}
 \frac{\zeta \left( 2 k + 1 \right)}{2^{2 k} \left( 2 k + 1 \right)} $$


UPDATE:
Thanks for your reply quid. I just came across this while doing some calculations with the zeta function. The calculations are a bit too long to be posted, but in short it derives from 
$$\zeta(s) = \frac{s+1}{2(s-1)} + \frac{s}{8} - \frac{s(s+1)}{2\pi^2}\int\_1^\infty \frac{(\tan^{-1}\cot(\pi x))^2}{x^{s+2}}dx$$.



Series representation for Euler-Mascheroni constant
In his 1887 paper *Table des valeurs des sommes* $S\_k = \sum\_{1}^\infty n^{-k}$ (Acta Mathematica 10 (1887), 299-302; volume available [online](http://archive.org/details/actamathematica24lefgoog)), Stieltjes used almost exactly this formula to compute Euler's constant to 33 decimal places. Of course as quid points out you need to know the zeta values to do this, but the main point of this paper was to compute those values, so he was just getting Euler's constant as a corollary. He uses a slight variant of the formula, with $\zeta(2k+1)-1$ in place of $\zeta(2k+1)$ for faster convergence (and a corresponding adjustment in the other term, which becomes $1+\log 2 - \log 3$). He derives the formula by taking the Taylor series expansion of $\log \Gamma(1+x)$ and using it to compute $\log \Gamma(1+1/2) - \log \Gamma(1-1/2)$.


