<s>[INST] Prove or disprove $n \geq 2 ~\rightarrow~ \prod \limits_{i=1}^{n} \left ( 1 - \frac{1}{i^2} \right ) ~=~ \frac{n+1}{2n}$ I am working on one of my HW assignments
$$
\forall n \in \mathbb{Z}, ~
 n \geq 2
 ~\rightarrow~
 \prod \limits_{i=1}^{n} \left (
  1 - \frac{1}{i^2}
 \right )
 ~=~
 \frac{n+1}{2n}
$$
And i am not clear whether it should be proved or disproved.
my main concern is base case n=2.
$$P_l (2) =  \prod \limits_{i=1}^{2} \left ( 1 - \frac{1}{i^2}  \right ) = \left ( 1 - \frac{1}{1^2}  \right ) \cdot \left ( 1 - \frac{1}{2^2} \right ) = 0 \cdot \frac{3}{4} = 0$$
$$P_r (2) =  \frac{2+1}{2 \cdot2} = \frac{3}{4}$$
but prove for k+1 works
Induction Hypothesis
$$  \prod \limits_{i=1}^{n} \left (
  1 - \frac{1}{i^2}
 \right )
 ~=~
 \frac{n+1}{2n} \rightarrow 
     \prod \limits_{i=1}^{n+1} \left (
  1 - \frac{1}{i^2}
 \right )
 ~=~
 \frac{(n+1)+1}{2(n+1)}= \frac{(n+2)}{2(n+1)}
$$
$$
 \prod \limits_{i=1}^{n+1} \left (
  1 - \frac{1}{i^2}
 \right )
 = \prod \limits_{i=1}^{n} \left (
  1 - \frac {1}{i^2}
 \right )\cdot \left (1 - \frac {1}{(n+1)^2} \right )
$$
By substitution
$$
 = \frac{n+1}{2n} \cdot \left (1 - \frac {1}{(n+1)^2} \right )
$$
$$
 = \frac{n+1}{2n} \cdot \frac{n(n+2)}{(n+1)^2}
$$
$$
 = \frac{n \cdot (n+1)\cdot (n+2)}{2n\cdot (n+1)\cdot (n+1)} = \frac{(n+2)}{2(n+1)}
$$
Now why my base case isn't working? if it should work for all n>2
Can someone point out what am I doing wrong?
 [/INST] \begin{align}
\prod \limits_{i=1}^{n} \Big(1 - \frac{1}{i^2}\Big)&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \prod \limits_{i=2}^{n} \Big(1 - \frac{1}{i^2}\Big)\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{\prod_{i=2}^{n}(i^2-1)}{\prod_{i=2}^{n}i^2}\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{\prod_{i=2}^{n}(i-1)\prod_{i=2}^{n}(i+1)}{(\prod_{i=2}^{n}i)^2}\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{(\frac1n\prod_{i=1}^{n}i)\Big(\frac{n+1}{2}\times\prod_{i=1}^{n}i\Big)}{(\prod_{i=1}^{n}i)^2}\\
&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{n+1}{2n}
\end{align}
</s>