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#1 h Apr 26, 2025, 9:43 AM

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Find the value of limit
$\Omega = \prod_{n=2}^{\infty} \frac{n^{14} - n^{12} - 2n^8 + 2n^6 + n^2 - 1}{n^{14}}$

This post has been edited 2 times. Last edited by P162008, Apr 26, 2025, 9:43 AM
Reason: Typo

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#2 h Apr 26, 2025, 9:37 PM • 1 Y

Y by P162008

Nice one. Factoring the numerator gives $\Omega=\Pi_2^\infty {{(n+1)^3}\over {n^3}}\cdot {{(n-1)^3}\over {n^3}}\cdot {{(n^2-n+1)^2(n^2+n+1)^2}\over {n^8}}$ .
Now $\Pi_2^n {{(k+1)^3(k-1)^3}\over {k^3\cdot k^3}}={{(n+1)^3}\over 8} \cdot {1\over {n^3}}=1/8(1+1/n)^3\rightarrow 1/8$ hence $\Omega=1/8\{ \Pi_2^\infty {{n^4+n^2+1}\over {n^4}}\}^2$ .
Write $\zeta:=\exp(2\pi i/3)$ then $x^2+x+1=(x-\zeta)(x-\overline{\zeta})$ hence $\Pi_2^\infty {{n^4+n^2+1}\over {n^4}}=1/3 \Pi_1^\infty (1-\zeta/n^2)\cdot \Pi_1^\infty (1-\overline{\zeta}/n^2)=1/3 |\Pi_1^\infty (1-\zeta/n^2)|^2$ .

Now we recall $\sin \pi z=\pi z\Pi_1^\infty (1-z^2/n^2)$ . Take $\alpha:=\exp(\pi i/3)$ then ${{\sin \pi \alpha}\over {\pi\alpha}}=\Pi_1^\infty (1-\zeta/n^2)$ .
Hence $|\Pi_1^\infty (1-\zeta/n^2)|^2={{|\sin \pi \alpha|^2}\over {\pi^2}}$ . Finally we use $|\sin z|^2=\sin^2 x+\sinh^2 y$ hence $|\sin \pi \alpha|^2=\sin^2(\pi/2)+\sinh^2 (\pi \sqrt{3}/2)=\cosh^2(\pi\sqrt{3}/2)$ .

This post has been edited 1 time. Last edited by alexheinis, Apr 26, 2025, 9:44 PM

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