Number 10, the infinite product is finite and equal to ##\pi / 2##. For this I wrote
##\prod_{n = 1 }^N \frac{4n^2}{4n^2-1} = 4^N (N!)^2 \frac{2^N N!}{(2N)!} \frac{2^N N!}{(2N+1)!} = \frac{4^{2N} (N!)^4}{(2N+1) ((2N)!)^2 } ##
And Stirling's formula leads to the conclusion