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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.

Note that for Harmonic numbers, $H_n$,
$$
sum_{k=1}^nleft(frac1k-logleft(1+frac1kright)right)=H_n-log(n+1)tag{1}
$$
Taking $(1)$ to the limit gives
$$
gamma=sum_{k=1}^inftyleft(frac1k-logleft(1+frac1kright)right)tag{2}
$$
We have the power series
$$
logleft(frac{1+x}{1-x}right)=2left(x+frac{x^3}{3}+frac{x^5}{5}+frac{x^7}{7}dotsright)tag{3}
$$
If we set $x=frac1{2k+1}$, then $frac{1+x}{1-x}=1+frac1k$; that is,
$$
logleft(1+frac1kright)=sum_{j=0}^inftyfrac2{2j+1}left(frac1{2k+1}right)^{2j+1}tag{4}
$$
Furthermore,
$$
begin{align}
sum_{k=1}^inftyleft(frac1{2k+1}right)^{2j+1}
&=sum_{k=1}^inftyleft(frac1{2k}right)^{2j+1}
+left(frac1{2k+1}right)^{2j+1}-frac1{2^{2j+1}}left(frac1{k}right)^{2j+1}\
&=zeta(2j+1)-1-frac1{2^{2j+1}}zeta(2j+1)\
&=frac{2^{2j+1}-1}{2^{2j+1}}(zeta(2j+1)-1)-frac1{2^{2j+1}}tag{5}
end{align}
$$
Then, using $(4)$ in $(2)$, and then applying $(5)$, we get
$$
begin{align}
gamma
&=sum_{k=1}^inftyleft(frac1k-logleft(1+frac1kright)right)\
&=sum_{k=1}^inftyfrac2{2k}-frac2{2k+1}-sum_{j=1}^inftyfrac2{2j+1}left(frac1{2k+1}right)^{2j+1}\
&=2(1-log(2))-sum_{j=1}^inftyfrac2{2j+1}left(frac{2^{2j+1}-1}{2^{2j+1}}(zeta(2j+1)-1)-frac1{2^{2j+1}}right)\
&=1-log(4)+log(3)-sum_{j=1}^inftyfrac{2^{2j+1}-1}{2^{2j}(2j+1)}(zeta(2j+1)-1)tag{6}
end{align}
$$
The last equality in $(6)$ follows from plugging $k=frac12$ into $(4)$ to get
$$
log(3)-1=sum_{j=1}^inftyfrac2{(2j+1)2^{2j+1}}tag{7}
$$
We can accelerate the convergence of $(6)$ once, using $(3)$, we compute
$$
begin{align}
sum_{j=1}^inftyfrac{zeta(2j+1)-1}{2j+1}
&=sum_{j=1}^inftysum_{k=2}^inftyfrac1{(2j+1)k^{2j+1}}\
&=lim_{ntoinfty}sum_{k=2}^nfrac12logleft(frac{k+1}{k-1}right)-frac1k\
&=lim_{ntoinfty}frac12log(n(n+1)/2)-(H_n-1)\
&=1-log(2)/2-gammatag{8}
end{align}
$$
If we add twice the left side of $(8)$ to the right side of $(6)$ and vice versa, we get
$$
2-log(2)-gamma
=1-log(4)+log(3)+sum_{j=1}^inftyfrac{zeta(2j+1)-1}{2^{2j}(2j+1)}tag{9}
$$
From which we get the Euler-Stieltjes series:
$$
gamma=1-log(3/2)-sum_{j=1}^inftyfrac{zeta(2j+1)-1}{4^j(2j+1)}tag{10}
$$

Using the following application of $(3)$
$$
sum_{j=1}^inftyfrac{frac1{n^{2j+1}}}{4^j(2j+1)}=logleft(frac{2n+1}{2n-1}right)-frac1ntag{11}
$$
we can accelerate the convergence of $(10)$:
$$
gamma=H_n-log(n+1/2)-sum_{j=1}^inftyfrac{zeta(2j+1)-sumlimits_{k=1}^nfrac1{k^{2j+1}}}{4^j(2j+1)}tag{12}
$$
$(12)$ converges about $2log_{10}(2n+2)$ digits per term. Euler-Stieltjes is the case $n=1$ of $(12)$. Note that $zeta(2j+1)-sumlimits_{k=1}^nfrac1{k^{2j+1}}=zeta(2j+1,n+1)$, the Hurwitz Zeta function.

In this answer, I give a another method for computing $gamma$ that uses the an accelerated function for the sum of the tail of the alternating harmonic series.

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:
$$gamma = frac{3}{2} - ln 2 - sum_{nge 2}frac{1}{n}left(zeta(n)-1- frac{1}{2^n}right)$$
$$gamma = frac{11}{6} - ln 3 - sum_{nge 2}frac{1}{n}left(zeta(n)-1
-frac{1}{2^n} -frac{1}{3^n}right)$$
$$gamma = 1- lnleft(frac{3}{2}right) -sum_{nge1}frac{zeta(2n+1)-1}{4^n(2n+1)} qquadtext{(Euler-Stieltjes)}
$$
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.

Do you mean this?
$$sum_{k=2}^infty {zeta(k)-1over k}= 1-gamma $$

This formula can be found in MathWorld (eq 123).

(Quoted in What is the fastest/most efficient algorithm for estimating Euler's Constant γ?.)