Product of Sequence of 1 minus Reciprocal of Squares

Theorem
For all $n \in \Z_{\ge 1}$:


 * $\displaystyle \prod_{j \mathop = 2}^n \left({1 - \dfrac 1 {j^2} }\right) = \dfrac {n + 1} {2 n}$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \prod_{j \mathop = 2}^n \left({1 - \dfrac 1 {j^2} }\right) = \dfrac {n + 1} {2 n}$

It is first noted that $n = 0$ is excluded because in that case $\dfrac {n + 1} {2 n}$ is undefined.

$P \left({1}\right)$ is the other edge case:

Thus $P \left({1}\right)$ is seen to hold.

Basis for the Induction
$P \left({2}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \prod_{j \mathop = 2}^k \left({1 - \dfrac 1 {j^2} }\right) = \dfrac {k + 1} {2 k}$

from which it is to be shown that:
 * $\displaystyle \prod_{j \mathop = 2}^{k + 1} \left({1 - \dfrac 1 {j^2} }\right) = \dfrac {k + 2} {2 \left({k + 1}\right)}$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{\ge 1}: \prod_{j \mathop = 2}^n \left({1 - \dfrac 1 {j^2} }\right) = \dfrac {n + 1} {2 n}$