# Proving Induction Step

For all $n>=2$, the formula $(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})\dots(1-\frac{1}{n^2}) = \frac{n+1}{2n}$

Proof:

Base case: $n = 2$.

$(1-\frac{1}{4}) = 0.75 = \frac{(2)+1}{2(2)}$

The claim holds for $n = 2.$

Inductive step: $n \geq 2$.

Suppose that $(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})\dots(1-\frac{1}{n^2}) = \frac{n+1}{2n}$ for $n \geq 2$

I want to show that $(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})\dots(1-\frac{1}{(n+1)^2}) = \frac{(n+1)+1}{2(n+1)}$

I've gotten it to the point of what I want to show but I don't really know how to do the math to prove this. Also, is there a better way to format fractions on this site? Because I find this a little difficult to read. Thanks for any help!