I am just learning about this, so I am a bit rough. I have come to a bit of a problem, and I cannot find out where I went wrong. My original question is:
$$\text{For every } n \ge 2, \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots \left(1-\frac{1}{n^2}\right) =\frac{n+1}{2n}$$
I went through this process, but I cannot find my mistake:
You've got the steps right, but it is a product.
You did $P(2)$. That working still holds.
We assume $P(k)$ holds:$$\prod_{i=1}^k(1-\frac{1}{i^2})=\frac{k+1}{2k}$$ And use it to show that $P(k+1)$ holds too.
$$\prod_{i=1}^{k+1}(1-\frac{1}{i^2})=(1-\frac{1}{(k+1)^2})\prod_{i=1}^k(1-\frac{1}{i^2})=(1-\frac{1}{(k+1)^2})(\frac{k+1}{2k})=\frac{(k^2+2k)(k+1)}{2k(k+1)^2}$$
Can you reduce this to $\frac{(k+1)+1}{2(k+1)}=\frac{k+2}{2k+2}$?