Evaluation of Euler's Constant $\gamma$

Question

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.

Answer 1

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

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Using the following application of $(3)$ $$ \sum_{j=1}^\infty\frac{\frac1{n^{2j+1}}}{4^j(2j+1)}=\log\left(\frac{2n+1}{2n-1}\right)-\frac1n\tag{11} $$ we can accelerate the convergence of $(10)$: $$ \gamma=H_n-\log(n+1/2)-\sum_{j=1}^\infty\frac{\zeta(2j+1)-\sum\limits_{k=1}^n\frac1{k^{2j+1}}}{4^j(2j+1)}\tag{12} $$ $(12)$ converges about $2\log_{10}(2n+2)$ digits per term. Euler-Stieltjes is the case $n=1$ of $(12)$. Note that $\zeta(2j+1)-\sum\limits_{k=1}^n\frac1{k^{2j+1}}=\zeta(2j+1,n+1)$, the Hurwitz Zeta function.

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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.

Answer 2

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.

Answer 3

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?.)

Answer 4

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 $|s|<1$

$$\ln\left(\Gamma\left(1-s\right)\right) = \gamma s + \sum_{n=2}^{\infty} \frac{\zeta\left(n\right)s^{n}}{n}$$