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I have recently come across this infinite product, and I was wondering what methods I could use to express the product in closed-form (if it is even possible):

$$\prod_{n=0}^{\infty}\dfrac{(4n+3)^{1/(4n+3)}}{(4n+5)^{1/(4n+5)}}=\dfrac{3^{1/3}}{5^{1/5}}\cdot \dfrac{7^{1/7}}{9^{1/9}}\cdot\dfrac{11^{1/11}}{13^{1/13}}\cdot\dfrac{15^{1/15}}{17^{1/17}}\cdot\dotsb$$

Thanks in advance!

StubbornAtom
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Harukr
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    Can you give the context? Many times, the subject surrounding it is a hint to where to start thinking. – Chrystomath Jul 23 '20 at 09:39
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    Taking the logarithm is usually a good start to convert the product to a sum. Then the powers can be brought down. – Benedict W. J. Irwin Jul 23 '20 at 09:39
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    Yes, it has a closed form. See https://math.stackexchange.com/questions/3032875/prove-sum-n-1-infty-frac-1n1-ln2n12n1-pi-4-gamma-ln-p – Robert Z Jul 23 '20 at 09:49
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    @Chrystomath I was attempting to solve the following: Choose $x$, $y$, and $z$ from $(0,1)$. Find the probability that $\frac{x}{yz}$ rounds to an even number. This infinite product appeared in a part of the final answer – Harukr Jul 24 '20 at 02:08

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$$\exp\sum_{n\geq 0}\left(\frac{\log(4n+3)}{4n+3}-\frac{\log(4n+5)}{4n+5}\right)=\exp\left[-\sum_{n\geq 1}\frac{\chi_4(n)\log(n)}{n}\right]$$ equals $$\lim_{s\to 1^+}\exp\left[\frac{d}{ds}\sum_{n\geq 1}\frac{\chi_4(n)}{n^s}\right]=\exp\beta'(1)$$ with $\beta$ being Dirichlet's beta function. According to equation (18) the closed form is $$ \exp\,\left[\frac{\pi}{4}\left(\gamma+\log\frac{4\pi^3}{\Gamma\left(\frac{1}{4}\right)^4}\right)\right].$$ A derivation can be found here.

Jack D'Aurizio
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