Prove that product from n=2 to infinity (1- 1/(n²))=1/2
Prove that $\displaystyle\prod_{n=2}^{\infty}\left(1-\frac{1}{n^2}\right)=\frac{1}{2}$.
Explanation of Method
Each factor $1-1/n^2$ splits into $\frac{n-1}n\cdot\frac{n+1}n$, turning the partial product into two independent telescoping products — one collapsing to $1/N$, the other to $(N+1)/2$ — whose product has a clean limit. The same result also falls out of Euler's product formula for $\sin(\pi x)$, evaluated near $x=1$.
- Factor $1-1/n^2=\dfrac{(n-1)(n+1)}{n^2}$ and split into two telescoping products.
- Evaluate each telescoping product in closed form for finite $N$.
- Take the limit $N\to\infty$, and cross-check via Euler's sine product.
Step 1 · Factor and split
$$ 1-\frac1{n^2} = \frac{(n-1)(n+1)}{n^2}, \qquad P_N=\prod_{n=2}^N\left(1-\frac1{n^2}\right) = \left(\prod_{n=2}^N\frac{n-1}n\right)\left(\prod_{n=2}^N\frac{n+1}n\right). $$
Step 2 · Two telescoping products
$$ \prod_{n=2}^N\frac{n-1}n = \frac12\cdot\frac23\cdot\frac34\cdots\frac{N-1}N = \frac1N, \qquad \prod_{n=2}^N\frac{n+1}n = \frac32\cdot\frac43\cdot\frac54\cdots\frac{N+1}N = \frac{N+1}2. $$ (Each is a standard telescoping product: every intermediate factor cancels between consecutive terms, leaving only the first denominator and last numerator.)
Step 3 · Combine and take the limit
$$ P_N = \frac1N\cdot\frac{N+1}2 = \frac{N+1}{2N}. $$ $$ \boxed{\lim_{N\to\infty}P_N = \lim_{N\to\infty}\frac{N+1}{2N} = \lim_{N\to\infty}\frac{1+1/N}2 = \frac12} $$ (Numerical check: $P_2=3/4$, $P_3=2/3$, $P_4=5/8$, $P_5=3/5$, $P_{10}=11/20=0.55$ — steadily decreasing toward $0.5$.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=\dfrac{N+1}{2N}$ |
| Limit | $\dfrac12$ |
Additional Notes
- The same value falls out of Euler's product $\sin(\pi x)=\pi x\prod_{n=1}^\infty(1-x^2/n^2)$: dividing out the $n=1$ factor gives $\prod_{n=2}^\infty(1-x^2/n^2)=\dfrac{\sin(\pi x)}{\pi x(1-x^2)}$ for $x\ne1$, and expanding both $\sin(\pi x)$ and $1-x^2$ to first order near $x=1$ (writing $x=1+\varepsilon$) gives $\dfrac{-\pi\varepsilon}{\pi(-2\varepsilon)}\to\dfrac12$ as $\varepsilon\to0$ — matching the telescoping result exactly.
- The factorial form from Method 2, $P_N=\dfrac{(N-1)!\,(N+1)!}{2\,(N!)^2}$, is a compact way to see the same cancellation: since $(N+1)!/N!=N+1$ and $(N-1)!/N!=1/N$, it collapses immediately to $\dfrac{N+1}{2N}$.
- This is a companion result to the classical Wallis-product family of telescoping/infinite-product identities; unlike the Wallis product for $\pi/2$, this one telescopes to a rational limit because the exponent is $2$ rather than requiring an alternating even/odd split.