Dtjytyk

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 $sumfrac1{n^2}=pi^2/6$ 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 $(4/7)^2approx1/3$, we can write
$$
sqrt{frac{1}{3}}=(frac{16}{48})^{1/2}
=(frac{16}{49}cdotfrac{49}{48})^{1/2}=frac{4}{7}(1+frac{1}{48})^{1/2}
$$
which we can expand as a binomial series, so $frac{4}{7}cdotfrac{97}{96}$ 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 $frac13-gamma^2$?

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
$$3gamma^2approx3left(frac{26}{45}right)^2=3frac{676}{2025}=3frac{676}{3cdot675}=frac{676}{675}approx 1$$

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,

$$ forall xin(0,1),qquad log(x)approxfrac{90(x-1)}{7(x+1)+12sqrt{x}+32sqrt{2x+(x+1)sqrt{x}}}tag{A} $$
and $approx$ holds as a $leq$, actually. We have
$$ gamma = int_{0}^{1}-log(-log x),dx tag{B}$$
hence:
$$ gammaleq 1-log(90)+int_{0}^{1}logleft[7(x+1)+12sqrt{x}+32sqrt{2x+(x+1)sqrt{x}}right],dxtag{C}$$
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.

http://forums.xkcd.com/viewtopic.php?f=17&t=107888#p3570666