<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] You are close.
You want to show that
$\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) 
= \frac{(k+1)+1}{2k+2}
$.
Note that the denominator on the right is
$2(k+1) = 2k+2$,
not the $2k+1$ you have.
That is why you were unable
to complete the proof -
you were trying to prove
something that was false.
$\begin{array}\\
\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right)
&=\frac {k+1}{2k} \times \left(\frac{(k+1)^2-1}{(k+1)^2}\right)\\
&=\frac {1}{2k} \times \left(\frac{k^2+2k+1-1}{k+1}\right)\\
&=\frac {1}{2k} \times \left(\frac{k^2+2k}{k+1}\right)\\
&=\frac {1}{2k} \times \left(\frac{k(k+2)}{k+1}\right)\\
&=\frac{k+2}{2(k+1)}\\
\end{array}
$
which is what you want.
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