<s>[INST] Proof by Induction $\frac{n+1}{2n} = \left(1-\frac{1}{2^2}\right)\dotsc \left(1-\frac{1}{n^2}\right)$ proof by induction. I'm pretty close to done with this proof but got stuck at the end 
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{n^2}\right) = \frac {n+1}{2n}$$
for all integers $n \geq 2$


*

*basis step
$$1-\frac{1}{2^2} = .75  \qquad \frac{2+1}{2\times2} = 3/4= .75$$

*set $n = k$ assumed to be true
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$

*set $n = k+1$
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1}$$
from step 2 we know that
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$
so we are going to plug that in
$$\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1}   $$ 
now to get both sides equal we multiply by $2k$ to get common denominators
$$\frac{k+1}{2k} \times {\left(1-{1\over(k+1)^2}\right) \times 2k \over 2k}$$   This is the left hand side
now to combine it 
$${\left((k+1) \left(1- {1\over(k+1)^2}\right) \times 2k \right)\over 2k}$$     - this is where I got stuck...
I wanted to cancel the $2k$'s on the top and bottom but then there would be no way to get it to equal the $(k+1)+1\over2(k+1)$ because the denominator required a $2k$. Any help is appreciated thank you.
 [/INST] Why not multiply the whole thing by $2k\cdot (k+1)^2\cdot 2(k+1)$ and see if you get an identity (after correcting the denominator of the RHS) ...
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