Product of Sequence of 1 minus Reciprocal of Squares/Proof 2

Theorem

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

$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$

We have:

$\ds \prod_{j \mathop = 2}^n \paren {1 - \frac 1 {j^2} }$

$=$

$\ds \prod_{j \mathop = 2}^n \paren {\frac {\paren {j - 1} \paren {j + 1} } {j^2} }$

$\ds $

$=$

$\ds \paren {\prod_{j \mathop = 2}^n \paren {j - 1} } \paren {\prod_{j \mathop = 2}^n \paren {j + 1} } \paren {\prod_{j \mathop = 2}^n \frac 1 j}^2$

$\ds $

$=$

$\ds \paren {\prod_{j \mathop = 1}^{n - 1} j} \paren {\prod_{j \mathop = 3}^{n + 1} j} \frac 1 {\paren {\prod_{j \mathop = 2}^n j}^2}$

$\ds $

$=$

$\ds \paren {n - 1}! \paren {\frac 1 2 \prod_{j \mathop = 2}^{n + 1} j} \frac 1 {\paren {n!}^2}$

Definition of Factorial

$\ds $

$=$

$\ds \frac {\paren {n - 1}! \paren {n + 1}!} {2 \times n! \times n!}$

Definition of Factorial

$\ds $

$=$

$\ds \frac {\paren {n - 1}! n! \paren {n + 1} } {2 \paren {n - 1}! \times n \times n!}$

$\ds $

$=$

$\ds \frac {n + 1} {2 n}$

$\blacksquare$