Euler Mascheroni Constant, Curious?

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 $$\zeta(s) = \frac{s+1}{2(s-1)} + \frac{s}{8} - \frac{s(s+1)}{2\pi^2}\int_1^\infty \frac{(\tan^{-1}\cot(\pi x))^2}{x^{s+2}}dx$$.

-
Could you please recheck the formula you state. I have no good access to a CAS at the moment but the first sum seems to be soomething like 0.693 and the latter 0.177 so the diff is quite a bit smaller than 0.57. – quid Aug 14 '12 at 12:08
I am using Mathematica, and "Sum[1./(k*2^k) - Zeta[2*k + 1]/((2 k + 1)*2^(2*k)), {k, 1, 100}]" gives me 0.5772156649015329 – Roupam Ghosh Aug 14 '12 at 12:46
Sorry for the preceeding comment. You are right. I used Wolfram Alpha and got the other values, but I now retried and got what you say. Although I tried to be careful, in all likelihood I just somehow made an error before; sorry again for the noise. – quid Aug 14 '12 at 12:59
@Roupam Do you have a reference or online resource for details about the integral representation of $\zeta(s)$ given above? – poirot Feb 4 '14 at 23:35
@pbs I posted this from my own work. I'm not sure if there's an online resource for it. – Roupam Ghosh Jan 1 '15 at 15:39

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), 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 $\log \Gamma(1+1/2) - \log \Gamma(1-1/2)$.

-
Wow... I will look at that paper for sure! :) Thanks. – Roupam Ghosh Aug 14 '12 at 12:47
Yep... pretty close to what I got... Here's the link for others to see... archive.org/stream/actamathematica24lefgoog#page/n308/mode/2up – Roupam Ghosh Aug 14 '12 at 14:31
For the records, here is the link for Legendre's calculation using Euler-Maclaurin summation: gallica.bnf.fr/ark:/12148/bpt6k1101484/f440.image, Traité des fonctions elliptiques et des intégrales eulériennes, Paris, (1825-1828), vol. 2, p. 434 – Papiro Aug 14 '12 at 19:23

The method used for the recent record computations of Euler--Mascheroni is (a refinement of) a classical algorithm due to Bren-McMillan . This algorithm is $O(n (\log n)^3)$.

Now, for you formula (btw, could you say where/how you found it?), it is not quite clear to me what you are asking.

The series involves $\zeta$ values (not only elementary things)! So, if one where to use this to compute approximations of $\gamma$ one would need all kinds of $\zeta$ values at odd naturals, and early ones to essentailly the same precision as one seeks $\gamma$. And, also from a theoretical this makes unclear what type of expressions you would admit as 'competition'.

To continue on the computation bit, if one would only compute/estimate $\zeta(3)$ naively this would already make this worse then above mentioned method. (One can also compute $\zeta(3)$ faster--in the recent record computations for it also an $O(n (\log n)^3)$ algorithm was used (though it appears to be simpler to compute $\zeta(3)$ than $\gamma$); but this is only $\zeta(3)$. And, it reraises the issue that it is unclear how to treat the $\zeta$ values when interpreting your question.)

An overview on algorithms used to calculate these and related constants to high precision in practise, can also be found on that site.

-
+1 For the links. I am just wanting to know whether this formula is unique and is of any use from a computational perspective. – Roupam Ghosh Aug 14 '12 at 10:50
You are welcome, and thank you for the added information. – quid Aug 14 '12 at 12:23