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 $\frac13-\gamma^2$?

## Best Answer

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: $$\gamma \approx \frac{7}{12}-\frac{1}{180}-\frac{1}{2310}=\frac{4001}{6930}<\frac{1}{\sqrt{3}}$$