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

Question

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.

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

Answer 1

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 $\frac{1}{\sqrt{3}}$.

Answer 2

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