What is the simplification of $ \frac{(1^4+1)(3^4+1)\cdots(19^4+1)}{(2^4+1)(4^4+1)\cdots(20^4+1)} $

Question

What is the simplification of $$ \frac{(1^4+1)(3^4+1)\cdots(19^4+1)}{(2^4+1)(4^4+1)\cdots(20^4+1)} $$ Can you help me simplify this expression I simpifed it and then I stuck here

Answer 1

Quoting @Jacky Chong, "It doesn't seem to have a nice form" seems to be an understatement.

Making the problem more general, we can get an approximation writing $$a_n=\frac{1^4+1}{2^4+1}\times\frac {\prod_{k=2}^n ((2k-1)^4+1) } {\prod_{k=2}^n ((2k)^4+1) }$$ The $1$ becomes quickly negligible compared to the fourth power. So, let us neglect it and write $$a_n \sim \frac{2}{17}\times\frac {\prod_{k=2}^n (2k-1)^4 } {\prod_{k=2}^n (2k)^4 }$$ Now $${\prod_{k=2}^n (2k-1)^4 } =\frac{16^n \Gamma \left(n+\frac{1}{2}\right)^4}{\pi ^2}\qquad\text{and}\qquad {\prod_{k=2}^n (2k)^4 }=16^{n-1} \Gamma (n+1)^4$$ So, $$a_n \sim \frac{32}{17 \pi ^2}\left(\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}\right)^4$$ For $n=10$ (this is your case) this would give $$a_{10}\sim \frac{267735104313263873}{147573952589676412928}=0.001814$$ instead of the rigorous $$a_{10}=\frac{31472190012985146835547604911104}{17183930189102530341505667303531209}=0.001831$$

For sure, we could make it better starting the product at $k=3$ and write $$a_n=\frac{164}{4369}\times\frac {\prod_{k=3}^n ((2k-1)^4+1) } {\prod_{k=3}^n ((2k)^4+1) }$$ making $$a_n \sim \frac{671744}{353889 \pi ^2} \left(\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}\right)^4$$ which, for $n=10$, would give $0.001829$.

Edit for your curiosity

Assuming that $n$ is large, we could use Stirling approximation for the factorial and get $$a_n \sim \frac{671744}{353889 \pi ^2 } \frac{e^{-\frac{1}{2 n}}}{n^2}$$ Applied for $n=10$, this would give $0.001829$ (surprising, ins't it ?).