Product of Sequence of 1 minus Reciprocal of Squares/Proof 1
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
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:
- $\ds \forall n \in \Z_{\ge 1}: \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$
$\blacksquare$