# How to separate <munderover> &#x220F;<!-- ∏ --> <mrow class="MJX-TeXAtom-ORD"> k

How to separate $\prod _{k=1}^{\mathrm{\infty }}\frac{\left(16k-15{\right)}^{\frac{1}{16k-15}}}{\left(16k-1{\right)}^{\frac{1}{16k-1}}}$ from Log-gamma function.
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Karla Hull
Let
$f\left(z\right)=\pi \left(\mathrm{ln}\mathrm{\Gamma }\left(z\right)+\left(z-\frac{1}{2}\right)\left(\gamma +\mathrm{ln}2\right)+\left(z-1\right)\mathrm{ln}\pi +\frac{1}{2}\mathrm{ln}\mathrm{sin}\left(\pi z\right)\right)=\sum _{k=1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(2\pi kz\right)}{k}\mathrm{ln}k$
Then assuming $\alpha \in \mathbb{N}$ and $\alpha >2$
$\mathrm{ln}\prod _{k=1}^{\mathrm{\infty }}\frac{\left(\alpha k-\alpha +1{\right)}^{\frac{1}{\alpha k-\left(\alpha -1\right)}}}{\left(\alpha k-1{\right)}^{\frac{1}{\alpha k-1}}}=\sum _{k=1}^{\mathrm{\infty }}\left(\frac{\mathrm{ln}\left(\alpha k-\alpha +1\right)}{\alpha k-\alpha +1}-\frac{\mathrm{ln}\left(\alpha k-1\right)}{\alpha k-1}\right)=\sum _{k=1}^{\mathrm{\infty }}s\left(k\right)\frac{\mathrm{ln}k}{k}$
$s\left(k\right)=\left\{\begin{array}{ll}1& k\equiv 1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}\alpha \right)\\ -1& k\equiv -1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}\alpha \right)\\ 0& \text{otherwise}\end{array}$
We can express s(k) as a sum of sin functions using the discrete fourier transform:
$s\left(k\right)=\frac{1}{\alpha }\sum _{n=1}^{\alpha -1}{s}_{n}\mathrm{sin}\left(\frac{2\pi nk}{\alpha }\right)$
where
${s}_{n}=\sum _{k=1}^{\alpha -1}s\left(k\right)\mathrm{sin}\left(\frac{2\pi nk}{\alpha }\right)=\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)-\mathrm{sin}\left(\frac{2\pi n\left(\alpha -1\right)}{\alpha }\right)=2\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)$
Hence,
$\sum _{k=1}^{\mathrm{\infty }}s\left(k\right)\frac{\mathrm{ln}k}{k}=\frac{2}{\alpha }\sum _{n=1}^{\alpha -1}\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)\sum _{k=1}^{\mathrm{\infty }}\mathrm{sin}\left(\frac{2\pi nk}{\alpha }\right)\frac{\mathrm{ln}k}{k}=\frac{2}{\alpha }\sum _{n=1}^{\alpha -1}\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)f\left(\frac{n}{\alpha }\right)$
Constant terms in f are annihilated in this formula, and
$\mathrm{sin}\left(2x\right)\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)=-\mathrm{sin}\left(2\left(\pi -x\right)\right)\mathrm{ln}\mathrm{sin}\left(\pi -x\right),$
so the lnsin terms are annihilated as well to give
$\frac{2\pi }{\alpha }\sum _{n=1}^{\alpha -1}\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)\left(\mathrm{ln}\mathrm{\Gamma }\left(\frac{n}{\alpha }\right)+\left(\frac{n}{\alpha }\right)\left(\gamma +\mathrm{ln}2\pi \right)\right)$
Using
$\sum _{n=1}^{\alpha -1}n\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)=-\frac{\alpha }{2}\mathrm{cot}\left(\frac{\pi }{\alpha }\right)$
which comes from taking the imaginary part of $\sum _{n=0}^{\alpha -1}n{z}^{n}=\frac{2z}{\left(1-z{\right)}^{2}}+\frac{\left(\alpha -1\right)z}{1-z}$ with $z=\mathrm{exp}\left(\frac{2\pi i}{\alpha }\right)$, we get
$\frac{2\pi }{\alpha }\sum _{n=1}^{\alpha -1}\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)\left(\frac{n}{\alpha }\right)\left(\gamma +\mathrm{ln}2\pi \right)=-\frac{\pi }{\alpha }\left(\gamma +\mathrm{ln}2\pi \right)\mathrm{cot}\left(\frac{\pi }{\alpha }\right)$
so we can finally write
$\prod _{k=1}^{\mathrm{\infty }}\frac{\left(\alpha k-\alpha +1{\right)}^{\frac{1}{\alpha k-\alpha +1}}}{\left(\alpha k-1{\right)}^{\frac{1}{\alpha k-1}}}={\left(2\pi {e}^{\gamma }\right)}^{-\frac{\pi }{\alpha }\mathrm{cot}\left(\frac{\pi }{\alpha }\right)}\prod _{n=1}^{\alpha -1}\mathrm{\Gamma }{\left(\frac{n}{\alpha }\right)}^{\frac{2\pi }{\alpha }\mathrm{sin}\left(\frac{2\pi n}{\alpha }\right)}$