Simple Division Problem I have the equation:
$$(1-\frac{1}{2^2})...(1-\frac{1}{n^2}) = \frac{n+1}{2n}$$ for n ≥ 2
Trying to prove by induction and I get the following equation.
$$\frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2} = \frac{k+2}{2(k+2)}$$ 
I can't to simplify it to the final answer.
I multiplied $$\frac{k+1}{2k}.{(k+1)^2}$$ and $$\frac{k(k+2)}{(k+1)^2}.{2k}$$ but I know something is wrong somewhere!
BTW, this is part of the proof by Induction step!

The statement is true for $n=2$. Suppose it holds for $n$; then
$$
\left(1-\frac{1}{2^2}\right)\dots
\left(1-\frac{1}{n^2}\right)
\left(1-\frac{1}{(n+1)^2}\right)=
\frac{n+1}{2n}\left(1-\frac{1}{(n+1)^2}\right)
$$
The final term becomes
$$
\frac{n+1}{2n}\frac{n(n+2)}{(n+1)^2}=\frac{(n+1)+1}{2(n+1)}
$$
which is exactly what was to be proved.
(Note: I prefer avoiding the change from $n$ to $k$, do as you like better.)
You had a $+$ in
$$
\frac{k+1}{2k} + \frac{k(k+2)}{(k+1)^2}
$$
where multiplication should be used.
