# Any integral or series to prove $\frac{1}{\sqrt{3}}>\gamma$? [duplicate]

I recently noticed that these two numbers are remarkably close:

$$\frac{1}{\sqrt{3}}-\gamma=0.000135\dots$$

Are there any integrals or series which can prove that $\frac{1}{\sqrt{3}}>\gamma$?

Meaning that (as usual in such cases) the function under the integral has to be non-negative and the value should be proportional to the difference of these numbers.

The same goes for the series (strictly non-negative terms).

A good overview for the inequalities with $\pi$, like $\frac{22}{7}>\pi$, can be found in this question.

A related question. But I don't consider my question a duplicate, because the linked question is more general.

• Possibly also helpful: math.stackexchange.com/questions/656283/… Commented Jun 2, 2016 at 22:25
• @Brenton, thank you. I see there are no 'pretty' answers so far Commented Jun 2, 2016 at 22:27
• I once tried with a series here math.stackexchange.com/a/1613229/134791 I bet the parenthesis is always negative, which is useful there, but I think I got stuck trying to prove it. Commented May 13, 2017 at 5:05

## 2 Answers

Interesting question. We may start with: $$\gamma = \sum_{n\geq 1}\frac{1}{n}\cdot[x^n]\frac{x}{\log(1-x)}=\sum_{n\geq 1}\frac{G_n}{n}. \tag{1}$$ Due to Steffensen's bounds, we know that the Gregory coefficients $[x^n]\frac{x}{\log(1-x)}$ behave like $\frac{1}{n\log^2 n}$, so we may try to apply the Cauchy-Schwarz inequality to $(1)$ in the trivial way, hoping to get a tight upper bound. That leads to:

$$\gamma \leq \sqrt{\zeta(2)\cdot\frac{1}{2\pi}\int_{0}^{2\pi}\left(\frac{e^{i\theta}}{\log(1-e^{i\theta})}+1\right)\left(\frac{e^{-i\theta}}{\log(1-e^{-i\theta})}+1\right)d\theta}$$ that simplifies to:

$$\gamma \leq \sqrt{\frac{\pi}{12}\left(2\pi+\int_{0}^{2\pi}\frac{d\theta}{\log(1-e^{i\theta})\log(1-e^{-i\theta})}\right)}\tag{2}$$ depending on an interesting integral, but just giving a weak upper bound.

Maybe computing the first $N$ terms of $(1)$, then applying the CS-approach above to $\sum_{n>N}(\ldots)$, it is possible to prove $\gamma<\frac{1}{\sqrt{3}}$ without the need to take a huge $N$. A simpler alternative is:

$$\gamma\leq \sum_{n=1}^{N}\frac{G_n}{n}+\sqrt{\left(\sum_{n>N}\frac{G_n}{n}\right)\left(\sum_{n>N}\frac{G_n}{n-1}\right)}\tag{3}$$ where the involved series can be computed in a explicit way in terms of rational numbers, $\log(\pi)$ and logarithms of natural numbers.

By taking $N=12$ in $(3)$ we get: $$\large\scriptstyle\gamma\leq \frac{198023355301039}{345226033152000}+\sqrt{\left(-\frac{28800521569}{50295168000}+\gamma\right) \left(-\frac{54074871014009}{86306508288000}-\frac{\gamma}{2}+\frac{\log(2\pi)}{2}\right)}\tag{4}$$ from which it follows that:

$$\gamma \leq \color{red}{\frac{-18276128754997+172613016576000 \log(2\pi)}{517839049728000}}\tag{5}$$

and the RHS of $(5)$ is less than $\frac{1}{\sqrt{3}}$.

• I'm still hoping for a 'prettier' answer, but thank you anyway, +1. I checked the Gregory series for $\gamma$ in Mathematica a while ago, and they seemed to have very slow convergence. Commented Jun 3, 2016 at 14:03
• @YuriyS: it converges like $\sum_{n\geq 2}\frac{1}{n^2\log^2 n}$, so yes, pretty slow, but acceleration techniques can be applied, due to the fact that $$\sum_{n\geq 1}\frac{G_n}{n+m}$$ can be computed in terms of logarithms due to Frullani's theorem. Commented Jun 3, 2016 at 14:31

I got a serie for you, but if you don't like inform me and I will delete it.

Let $$A=\left(1+{1\over \pi} \right)\left(6-\ln{27 }\right)-{\pi \over \sqrt3}$$ Then we have: $$\gamma+{1\over \sqrt3}=A+\sum_{n=1}^{\infty}\left({2+{2\over \pi} \over n+{1\over 3}}-{1+{2\over \pi} \over n}-\ln{n+1\over n}\right)$$

• It's not about liking or not liking: I find it hard to see how this anwers the question, since you provide the series for the sum, not the difference... In the end, the community decides if the answer is good or not by voting or commenting Commented May 22, 2017 at 11:56