Product of Sequence of 1 minus Reciprocal of Squares

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Theorem

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

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


Proof 1

The proof proceeds by induction.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:

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


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


$\map P 1$ is the other edge case:

\(\ds \prod_{j \mathop = 2}^1 \paren {1 - \dfrac 1 {j^2} }\) \(=\) \(\ds 1\) Definition of Vacuous Product
\(\ds \) \(=\) \(\ds \dfrac {1 + 1} {2 \times 1}\)

Thus $\map P 1$ is seen to hold.


Basis for the Induction

$\map P 2$ is the case:

\(\ds \prod_{j \mathop = 2}^2 \paren {1 - \dfrac 1 {j^2} }\) \(=\) \(\ds 1 - \dfrac 1 {2^2}\) Definition of Continued Product
\(\ds \) \(=\) \(\ds \frac {4 - 1} 4\)
\(\ds \) \(=\) \(\ds \frac {2 + 1} {2 \times 2}\)
\(\ds \) \(=\) \(\ds \valueat {\frac {n + 1} {2 n} } {n \mathop = 2}\)

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

\(\ds \prod_{j \mathop = 2}^{k + 1} \paren {1 - \dfrac 1 {j^2} }\) \(=\) \(\ds \prod_{j \mathop = 2}^k \paren {1 - \dfrac 1 {j^2} } \paren {1 - \dfrac 1 {\paren {k + 1}^2} }\) Definition of Continued Product
\(\ds \) \(=\) \(\ds \dfrac {k + 1} {2 k} \paren {1 - \dfrac 1 {\paren {k + 1}^2} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \dfrac {k + 1} {2 k} \times \dfrac {\paren {k + 1}^2 - 1} {\paren {k + 1}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {k^2 + 2 k + 1 - 1} {2 k \paren {k + 1} }\) simplification
\(\ds \) \(=\) \(\ds \dfrac {k \paren {k + 2} } {2 k \paren {k + 1} }\) further simplification
\(\ds \) \(=\) \(\ds \dfrac {k + 2} {2 \paren {k + 1} }\) further simplification


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


Therefore:

$\ds \forall n \in \Z_{\ge 1}: \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$

$\blacksquare$


Proof 2

We have:

\(\ds \prod_{j \mathop = 2}^n \paren {1 - \frac 1 {j^2} }\) \(=\) \(\ds \prod_{j \mathop = 2}^n \paren {\frac {\paren {j - 1} \paren {j + 1} } {j^2} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \paren {\prod_{j \mathop = 2}^n \paren {j - 1} } \paren {\prod_{j \mathop = 2}^n \paren {j + 1} } \paren {\prod_{j \mathop = 2}^n \frac 1 j}^2\) Product of Products
\(\ds \) \(=\) \(\ds \paren {\prod_{j \mathop = 1}^{n - 1} j} \paren {\prod_{j \mathop = 3}^{n + 1} j} \frac 1 {\paren {\prod_{j \mathop = 2}^n j}^2}\)
\(\ds \) \(=\) \(\ds \paren {n - 1}! \paren {\frac 1 2 \prod_{j \mathop = 2}^{n + 1} j} \frac 1 {\paren {n!}^2}\) Definition of Factorial
\(\ds \) \(=\) \(\ds \frac {\paren {n - 1}! \paren {n + 1}!} {2 \times n! \times n!}\) Definition of Factorial
\(\ds \) \(=\) \(\ds \frac {\paren {n - 1}! n! \paren {n + 1} } {2 \paren {n - 1}! \times n \times n!}\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \frac {n + 1} {2 n}\)

$\blacksquare$


Sources

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