I have found an infinite sum $$\prod_{n=1}^{\infty}\Big(1-\frac{(n+1)^2x^2}{n^4}\Big)$$ This function acts similarly to the sine function, as shown in the diagram below:
This function diverges when the value $n\rightarrow \infty$.
So the question is: what is the function that could represent this infinite product?
This function has its zeros when $x=\pm \frac{k^2}{k+1}$ where $k$ is any positive integer.
Using Wolfram Alpha, we do not get much except the interesting last expression using Pochammer symbols $$P_m=\prod_{n=1}^{m}\Bigg(1-\frac{(n+1)^2x^2}{n^4}\Bigg)$$ $$(m!)^4 P_m=\left(\frac{1}{2} \left(x-\sqrt{x-4} \sqrt{x}+2\right)\right)_m \times\left(\frac{1}{2} \left(x+\sqrt{x-4} \sqrt{x}+2\right)\right)_m \times$$ $$\left(\frac{1}{2} \left(-x-\sqrt{x-4} \sqrt{x}+2\right)\right)_m \times\left(\frac{1}{2} \left(-x+\sqrt{x-4} \sqrt{x}+2\right)\right)_m $$
Now, using asymptotics $$\prod_{n=1}^{\infty}\Bigg(1-\frac{(n+1)^2x^2}{n^4}\Bigg)$$ is the reciprocal of the product of four gamma functions $$\Gamma \left(\frac{1}{2} \left(x-\sqrt{x-4} \sqrt{x}+2\right)\right)\times \Gamma \left(\frac{1}{2} \left(x+\sqrt{x-4} \sqrt{x}+2\right)\right)\times $$ $$\Gamma \left(\frac{1}{2} \left(-x-\sqrt{x+4} \sqrt{x}+2\right)\right)\times \Gamma \left(\frac{1}{2} \left(-x+\sqrt{x+4} \sqrt{x}+2\right)\right)$$