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Calculate in closed form of:

$$\frac{\sqrt[5]{5}}{\sqrt[3]{3}}\cdot \frac{\sqrt[9]{9}}{\sqrt[7]{7}}\cdot \frac{\sqrt[13]{13}}{\sqrt[11]{11}}\cdot …$$

it has closed form. My solution uses the Kummer formula, and I would like to know if someone is encouraged to give a different solution

I obtain $\exp \left( -\frac{\pi \gamma }{4} \right)\cdot \left( \frac{\Gamma \left( \frac{1}{4} \right)}{\sqrt[4]{4\pi ^{3}}} \right)^{\pi }$

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The given infinite product equals

$$ P=\exp\sum_{n\geq 1}\left(\frac{\log(4n+1)}{4n+1}-\frac{\log(4n-1)}{4n-1}\right)=\exp\sum_{n\geq 1}\frac{\chi(n)\log(n)}{n}. $$

with $\chi$ being the non-principal Dirichlet’s character $\!\!\pmod{4}$.

In compact form, $P=\exp\left(-L'(\chi,1)\right)$. By Frullani’s Theorem $\log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx$, hence:

$$ P=\exp\left(-L'(\chi,1)\right) = \exp\int_{0}^{+\infty}\left(\frac{\pi}{4}e^{-x}-\arctan e^{-x}\right)\frac{dx}{x} $$

By considering the Taylor series of $\frac{\pi(1-z)}{4}-\arctan(1-z)$ around the origin we may easily derive the approximation $\color{blue}{P\approx 0.82}$. The previous integral representation recalls Binet’s second $\log\Gamma$ formula. Indeed, by the Malmsten-Kummer Fourier series, for any $z\in(0,1)$ we have:

$$ \log\Gamma(z) = \left(\tfrac12 – z\right)(\gamma + \log 2) + (1 – z)\ln\pi

– \frac12\log\sin\pi z + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\sin 2\pi n z \cdot\log n} n$$

hence an explicit representation for $P$ can be simply found by plugging in $z=\frac{1}{4}$:

$$\boxed{ P = \color{blue}{\left(\frac{\Gamma\left(\frac{1}{4}\right)^4}{4\pi^3 e^{\gamma}}\right)^{\frac{\pi}{4}}}\approx 0.82456334.}$$

For every $n\ge 3$ we have $$\sqrt[n]{n}>\sqrt[n+2]{n+2}$$ because it is equivalent to $n^2>(1+\frac{2}{n})^n$ which is true by comparison to the RHS limit: $n^2>e^2>(1+\frac{2}{n})^n$.

This proves that every term of your product is less then 1.

Obviously is also positive.

Is it enough to prove that the product equals 0?

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