Evaluation of Euler's Constant $\gamma$ Long back I had seen (in some obscure book) a formula to calculate the value of Euler's constant $\gamma$ based on a table of values of Riemann zeta function $\zeta(s)$. I am not able to recall the formula, but it used the fact that $\zeta(s) \to 1$ as $s \to \infty$ very fast and used terms of the form $\zeta(s) - 1$ for odd values of $s > 1$ (something like a series $\sum(\zeta(s) - 1)$). If anyone has access to this formula please let me know and it would be great to have a proof.
 A: There are a lot of formulas of this type. Some of them are in the Collection of formulae for Euler's constant $\gamma\;$ by Xavier Gourdon and Pascal Sebah:
$$\gamma = \frac{3}{2} - \ln 2 - \sum_{n\ge 2}\frac{1}{n}\left(\zeta(n)-1- \frac{1}{2^n}\right)$$
$$\gamma = \frac{11}{6} - \ln 3 - \sum_{n\ge 2}\frac{1}{n}\left(\zeta(n)-1 
-\frac{1}{2^n} -\frac{1}{3^n}\right)$$
$$\gamma = 1- \ln\left(\frac{3}{2}\right) -\sum_{n\ge1}\frac{\zeta(2n+1)-1}{4^n(2n+1)} \qquad\text{(Euler-Stieltjes)}
$$
The first two are derived from the Hurwitz zeta function as special cases. The Euler-Stieltjes formula seems near to your remembrance but is listed without proof.
Edit: You can find a proof in the Expansion of Euler's constant in terms of zeta numbers by M. Prévost.
A: Do you mean this?
$$\sum_{k=2}^\infty {\zeta(k)-1\over k}=  1-\gamma $$ 
This formula can be found in MathWorld (eq 123).
(Quoted in What is the fastest/most efficient algorithm for estimating Euler's Constant γ?.)
A: Theres quite a nice simple family of formulas that matches your description I'm aware of, based off the Taylor series of $ln\left(\Gamma\left(s\right)\right)$. However I believe this is only valid when $|s|<1$
$$\ln\left(\Gamma\left(1-s\right)\right) = \gamma s + \sum_{n=2}^{\infty} \frac{\zeta\left(n\right)s^{n}}{n}$$
