How do you evaluate $$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right)$$ using the identity $$\sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)?$$

I assume I'll have to express $1+\frac{1}{n^2}+\frac{1}{n^4}$ as $\left(1+\frac{a}{n^2}\right)\left(1+\frac{b}{n^2}\right)$ for some $a,b\in\mathbb{C}$ so that I could use the identity given, but I can't seem to factor the above appropriately.

  • $\begingroup$ I checked wolframalpha and it is indeed convergent with limit 5.89611... $\endgroup$ – Dante May 17 '17 at 7:33
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    $\begingroup$ In your identity, you can choose $a = (1+i\sqrt{3})/2$, $b = (1-i\sqrt{3})/2$. $\endgroup$ – Rigel May 17 '17 at 7:35
  • $\begingroup$ @Rigel WOW it worked. How on earth did you figure that factorization out? $\endgroup$ – Dante May 17 '17 at 7:38
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    $\begingroup$ Once you write $1+\frac{1}{n^2}+\frac{1}{n^4} = (1+\frac{a}{n^2})(1+\frac{b}{n^2})$, you see that you must have $a+b=1$, $ab = 1$ and then you solve this system (or you observe that $a$, $b$ are the solutions of the equation $z^2-z+1=0$). $\endgroup$ – Rigel May 17 '17 at 7:42
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    $\begingroup$ @Rigel I advise you to group your comments into an answer. I will surely upvote it. $\endgroup$ – Jean Marie May 17 '17 at 8:02

We have $1+z+z^2=\Phi_3(z)=(z-\omega)(z-\bar{\omega})=(1-\omega z)(1-\bar{\omega}z)$, with $\omega=\exp{\frac{2\pi i}{3}}$.
By considering $z=\frac{1}{n^2}$ and $\eta=\exp\frac{\pi i}{3}$ we have that $$ \frac{\sin(\pi w)}{\pi w}=\prod_{n\geq 1}\left(1-\frac{w^2}{n^2}\right)\tag{1} $$ implies: $$ \prod_{n\geq 1}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right) = \frac{\sin(\pi\eta)\sin(\pi\bar{\eta})}{\pi^2}=\frac{\cos(\pi(\eta-\bar{\eta}))-\cos(\pi(\eta+\bar{\eta}))}{2\pi^2}\tag{2}$$ so that: $$ \prod_{n\geq 1}\left(1+\frac{1}{n^2}+\frac{1}{n^4}\right) = \color{red}{\frac{1+\cosh(\pi\sqrt{3})}{2\pi^2}}=\left(\frac{1}{\pi}\,\cosh\frac{\pi\sqrt{3}}{2}\right)^2.\tag{3}$$


Observing that $$ \left(1+\frac1{n^2}+\frac1{n^4}\right) =\left(1-\frac{a^2}{n^2}\right)\left(1-\frac{b^2}{n^2}\right) $$ where $$ a^2+b^2=-1\\ a^2b^2=1 $$ find $a$ and $b$ by a direct computation.

We get

\begin{align*} \prod_{n\ge1}\left(1+\frac1{n^2}+\frac1{n^4}\right) =&\prod_{n\ge1}\left(1-\frac{a^2}{n^2}\right)\left(1-\frac{b^2}{n^2}\right)\\ =&\prod_{n\ge1}\left(1-\frac{a^2}{n^2}\right) \prod_{n\ge1}\left(1-\frac{b^2}{n^2}\right)\\ =&\frac{\sin(\pi a)}{\pi a}\frac{\sin(\pi b)}{\pi b} \end{align*}


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