# A product sequence

Evaluate $$\frac{\prod_{i=1}^n[(2i-1)^4+ 1/4]}{\prod_{i=1}^n[(2i)^4+ 1/4]}$$

First I thought I would multiply both the numerator and the denominator by the denominator itself. Now, I am unable to evaluate the series. I would appreciate innovative ideas to evaluate this product.

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We shall slightly rewrite the task: $$\frac{\prod_{i=1}^n[(2i-1)^4+ 1/4]}{\prod_{i=1}^n[(2i)^4+ 1/4]} = \prod_{i=1}^n\frac{[(2i-1)^4+ 1/4]}{[(2i)^4+ 1/4]}$$
First note the equality $$x^4 + \frac14 = (x^2 + x + \frac12)(x^2 - x + \frac12)$$ So a term in the product gives $$\frac{(2i-1)^4+ 1/4}{(2i)^4+ 1/4} = \frac{[(2i-1)^2+(2i-1)+ 1/2][(2i-1)^2-(2i-1)+ 1/2]}{[(2i)^2+2i+ 1/2][(2i)^2-2i+ 1/2]}$$ Now we multiply out and get $$(2i-1)^2+(2i-1)+\dfrac{1}{2}=(2i)^2-4i+1+2i-1+\dfrac{1}{2}=(2i)^2-(2i)+\dfrac{1}{2}$$ which makes the term in the product $$\frac{(2i-1)^4+ 1/4}{(2i)^4+ 1/4} = \frac{(2i-1)^2-(2i-1)+ 1/2}{(2i)^2+2i+ 1/2}$$ Trying to reduce the numerator from $$(2i-1)$$ to $$(2i-2)$$ we observe that $$(2i-2)^2+(2i-2)+\frac{1}{2}=(2i)^2-8i+4+2i-2+\frac{1}{2}=(2i-1)^2-(2i-1)+\frac{1}{2}$$ which makes the term in the product $$\frac{(2i-1)^4+ 1/4}{(2i)^4+ 1/4} = \frac{(2i-2)^2+(2i-2)+\frac{1}{2}}{(2i)^2+2i+ 1/2}= \frac{(2(i-1))^2+2(i-1)+\frac{1}{2}}{(2i)^2+2i+ 1/2}$$ Now we can perform the product, since by telescoping all "inner" numerators and denominators cancel: $$\prod_{i=1}^n\frac{[(2i-1)^4+ 1/4]}{[(2i)^4+ 1/4]} = \prod_{i=1}^n \frac{(2(i-1))^2+2(i-1)+\frac{1}{2}}{(2i)^2+2i+ 1/2} \\ = \frac{(2\cdot 1-2)^2+(2\cdot 1-2)+\frac{1}{2}}{(2\cdot n)^2+2\cdot n+ 1/2} = \frac{1}{8 n^2+ 4 n+ 1}$$ This completes the evaluation. $$\qquad \Box$$