# 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.

• The following paper has appeared just a few weeks ago: Jeffrey C. Lagarias: Euler's constant – Euler's work and modern developments. Bull. Am. Math. Soc. 50 (2013), 527–628. In this paper Lagarias has done for Euler's constant what Melville did for the whale. – Christian Blatter Nov 12 '13 at 14:27
• Only a note (off-topic). The series with value of $\zeta(n)$ are not the best way to evaluate $\gamma$. The best two easy ways to calculate Euler's constant numerically are:<br> (1) Euler-Maclaurin sumformula<br> (2) With an asymptotic expansion of li(x) [integrallogarithm]. This Method is going back to Heinrich Wilhelm Brandes (1777–1834) in 1824. – skraemer Mar 11 at 19:17

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: $$\gamma=1-\log(3/2)-\sum_{j=1}^\infty\frac{\zeta(2j+1)-1}{4^j(2j+1)}\tag{10}$$

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.

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.

• To robjohn: Thanks for the nice formula which is almost similar to Euler-Stieltjes. And the best part is the very very elementary proof which is accessible to anyone who knows the definition of $\gamma$ and the logarithmic series. I also checked your other answer and found it very informative. Thanks a lot (+1)! I have not yet accepted any answer to check for any more interesting answers. – Paramanand Singh Nov 12 '13 at 14:33
• To Robjohn: Before I could connect the dots from your answer to Euler-Stieltjes, you gave the full derivation.. gr8! – Paramanand Singh Nov 12 '13 at 14:43
• @ParamanandSingh: I have added a faster converging extension of Euler-Stieltjes. – robjohn Nov 12 '13 at 16:11
• To Robjohn: I am speechless! – Paramanand Singh Nov 12 '13 at 16:30

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.

• Yes I also think that the third one is what I saw long back. I will try to go through the linked PDF references – Paramanand Singh Nov 12 '13 at 10:05
• The first two series are neat in that they are the series given by lhf with the first two or three terms of $\zeta(n)$ taken out. They give better convergence at about $0.477$ and $.602$ places per term. The Euler-Stieltjes Series gives approximately $1.2$ places per term. (+1) – robjohn Nov 12 '13 at 15:26
• In the spirit of your first two series (which extend the series given by lhf), I have similarly accelerated the Euler-Stieltjes series in my answer. – robjohn Nov 12 '13 at 16:39

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

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

• No! the sum I had seen is mentioned in answer by gammatester namely $$\gamma = 1- \ln\left(\frac{3}{2}\right) -\sum_{n\ge1}\frac{\zeta(2n+1)-1}{4^n(2n+1)} \qquad\text{(Euler-Stieltjes)}$$ The formula which you mention has an easy proof. – Paramanand Singh Nov 12 '13 at 10:44
• Although it may not be the series that Paramanand was seeking, it is still a valid series, giving about $0.3$ places each term. (+1) – robjohn Nov 12 '13 at 15:20