Questions that ask for "intuitive" reasons are admittedly subjective, but I suspect some people will find this interesting.

Some time ago, I was struck by the coincidence that the Euler-Mascheroni constant $\gamma$ is close to the square root of $1/3$. (Their numerical values are about $0.57722$ and $0.57735$ respectively.)

Is there any informal or intuitive reason for this? For example, can we find a series converging to $\gamma$ and a series converging to $\sqrt{1/3}$ whose terms are close to each other?

An example of the kind of argument I have in mind can be found in Noam Elkies' list of one-page papers, where he gives a "reason" that $\pi$ is slightly less than $\sqrt{10}$. (Essentially, take $\sum\frac1{n^2}=\pi^2/6$ as known, and then bound that series above by a telescoping series whose sum is $10/6$.)

There are lots of ways to get series that converge quickly to $\sqrt{1/3}$. For example, taking advantage of the fact that $(4/7)^2\approx1/3$, we can write $$ \sqrt{\frac{1}{3}}=(\frac{16}{48})^{1/2} =(\frac{16}{49}\cdot\frac{49}{48})^{1/2}=\frac{4}{7}(1+\frac{1}{48})^{1/2} $$ which we can expand as a binomial series, so $\frac{4}{7}\cdot\frac{97}{96}$ is an example of a good approximation to $\sqrt{1/3}$. Can we also get good approximations to $\gamma$ by using series that converge quickly, and can we find the "right" pair of series that shows "why" $\gamma$ is slightly less than $\sqrt{1/3}$?

Another type of argument that's out there, showing "why" $\pi$ is slightly less than $22/7$, involves a particular definite integral of a "small" function that evaluates to $\frac{22}{7}-\pi$. So, are there any definite integrals of "small" functions that evaluate to $\sqrt{\frac13}-\gamma$ or $\frac13-\gamma^2$?

  • 3
    $\begingroup$ For the same reason that $2\pi+e$ is so clsoe to $9$. :-) $\endgroup$
    – Lucian
    Jan 29, 2014 at 18:38
  • 4
    $\begingroup$ Also, this link might prove helpful. $\endgroup$
    – Lucian
    Jan 29, 2014 at 18:44
  • $\begingroup$ Thank you for the link. Unfortunately, it's short on details, and I still don't fully understand how to conclude that $\gamma\approx\sqrt{1/3}$. I assume the Gaussian quadrature mentioned is the one that approximates $\int_{-1}^1 f(x)dx$ with $f(\sqrt{1/3})+f(-\sqrt{1/3})$ (rescaled as appropriate). I can use these ideas to approximate $\gamma$ with a series, but not one that is obviously close to $\sqrt{1/3}$. $\endgroup$
    – idmercer
    Jan 30, 2014 at 21:09
  • $\begingroup$ @Lucian For attempts at why $2\pi+e$ is so close to $9$ please visit math.stackexchange.com/questions/1711437/… $\endgroup$ May 13, 2017 at 5:14

3 Answers 3


From the continuous fraction expansion, the seventh convergent is

$$\gamma \approx \frac{15}{26}$$

From the limit definition

$$\begin{align} \gamma &= \lim_{n \to \infty} {\left(2H_n-\frac{1}{6}H_{n^2+n-1}-\frac{5}{6}H_{n^2+n}\right)} \\ &= \frac{7}{12}+\sum_{n=1}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\ &=\frac{7}{12}-\frac{1}{180}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\ &=\frac{26}{45}+\sum_{n=2}^{\infty}\left(\frac{2}{n+1}-\frac{1}{3n(n+1)(n+2)}-\sum_{k=n(n+1)+1}^{(n+1)(n+2)}\frac{1}{k}\right) \\ \end{align} $$

so $$\gamma \approx \frac{26}{45}$$

Multiplying both approximations,

$$\gamma^2 \approx \frac{1}{3}$$

The origin of this limit definition is improving the convergence of Macys formula from $o(n^{-2})$ to $o(n^{-4})$ downweighting the last term, without additional fractions (https://math.stackexchange.com/a/129808/134791)

In fact, the convergent approximation is not necessary.

Given $$\gamma \approx \frac{26}{45}$$

we have $$3\gamma^2\approx3\left(\frac{26}{45}\right)^2=3\frac{676}{2025}=3\frac{676}{3\cdot675}=\frac{676}{675}\approx 1$$

which also yields $$\gamma^2 \approx \frac{1}{3}$$

Towards proving that $\gamma < \frac{1}{ \sqrt {3} }$, we may take one more term out of the summation: $$\gamma \approx \frac{7}{12}-\frac{1}{180}-\frac{1}{2310}=\frac{4001}{6930}<\frac{1}{\sqrt{3}}$$


I have an interesting approach. The Shafer-Fink inequality and its generalization allow to devise algebraic approximations of the arctangent function with an arbitrary uniform accuracy. By a change of variable, the same holds for the hyperbolic arctangent function over the interval $(0,1)$ and for the logarithm function over the same interval. For instance,

$$ \forall x\in(0,1),\qquad \log(x)\approx\frac{90(x-1)}{7(x+1)+12\sqrt{x}+32\sqrt{2x+(x+1)\sqrt{x}}}\tag{A} $$ and $\approx$ holds as a $\leq$, actually. We have $$ \gamma = \int_{0}^{1}-\log(-\log x)\,dx \tag{B}$$ hence: $$ \gamma\leq 1-\log(90)+\int_{0}^{1}\log\left[7(x+1)+12\sqrt{x}+32\sqrt{2x+(x+1)\sqrt{x}}\right]\,dx\tag{C}$$ where the RHS of $(C)$ just depends on the (logarithms of the) roots of $7 + 32 x + 12 x^2 + 32 x^3 + 7 x^4$, which is a quartic and palindromic polynomial.
The numerical approximation produced by $(C)$ allows to state: $$ \gamma < 0.5773534 < \frac{\pi}{2e}.\tag{D}$$ Actually $(A)$ is not powerful enough to prove $\gamma<\frac{1}{\sqrt{3}}$, but we can achieve that too by replacing $(A)$ with the higher-order (generalized) Shafer-Fink approximation.


I also asked this question at the xkcd messageboards and got an answer that was enlightening to me.


  • 2
    $\begingroup$ The link does not work $\endgroup$
    – Anixx
    Apr 18, 2021 at 12:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .