Prove that product from n=3 to infinity
Prove that $\displaystyle\prod_{n=3}^{\infty}\frac{n^2-4}{n^2-1}=\frac14$.
Explanation of Method
Factoring $n^2-4=(n-2)(n+2)$ and $n^2-1=(n-1)(n+1)$ splits each factor into two shifted telescoping products, exactly as in Problem 360 but with a shift of $1$ in the indices. Each half-product collapses to a simple closed form in $N$, and the two combine to a clean limit.
- Factor $n^2-4$ and $n^2-1$, and split the partial product into two telescoping halves.
- Evaluate each telescoping product in closed form for finite $N$.
- Combine and take the limit $N\to\infty$.
Step 1 · Factor and split
$$ \frac{n^2-4}{n^2-1}=\frac{(n-2)(n+2)}{(n-1)(n+1)}, \qquad P_N=\prod_{n=3}^N\frac{n^2-4}{n^2-1}=\left(\prod_{n=3}^N\frac{n-2}{n-1}\right)\left(\prod_{n=3}^N\frac{n+2}{n+1}\right). $$
Step 2 · Two telescoping products
$$ \prod_{n=3}^N\frac{n-2}{n-1}=\frac12\cdot\frac23\cdot\frac34\cdots\frac{N-2}{N-1}=\frac1{N-1}, \qquad \prod_{n=3}^N\frac{n+2}{n+1}=\frac54\cdot\frac65\cdot\frac76\cdots\frac{N+2}{N+1}=\frac{N+2}4. $$
Step 3 · Combine and take the limit
$$ P_N=\frac1{N-1}\cdot\frac{N+2}4=\frac{N+2}{4(N-1)}. $$ $$ \boxed{\lim_{N\to\infty}P_N=\lim_{N\to\infty}\frac{1+2/N}{4(1-1/N)}=\frac14} $$ (Numerical check: $P_3=7/15\approx0.467$, $P_5\approx0.313$, $P_{10}\approx0.281$, $P_{1000}\approx0.2503$ — steadily approaching $0.25$; direct product to $10^5$ terms gives $0.250004$.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=\dfrac{N+2}{4(N-1)}$ |
| Limit | $\dfrac14$ |
Additional Notes
- This is the same telescoping mechanism as Problem 360's $\prod(1-1/n^2)=1/2$, but shifted: factoring $n^2-4=(n-2)(n+2)$ instead of $(n-1)(n+1)$ shifts the two half-products by an extra step, changing the limit from $1/2$ to $1/4$.
- The factorial form $P_N=\dfrac{(N-2)!\,(N+2)!/24}{(N-1)!\,(N+1)!/6}=\dfrac{(N+2)!\,(N-2)!}{4(N+1)!\,(N-1)!}$ collapses immediately to $\dfrac{N+2}{4(N-1)}$ using $(N+1)!/N!=N+1$, matching Method 1 exactly.
- General pattern: $\prod_{n=k+1}^\infty\dfrac{n^2-k^2}{n^2-1}=\dfrac1{2k}\cdot\dfrac{(k+1)}{1}\cdot\dfrac1{(k-1)!}$-type shifted telescoping identities all reduce to a ratio of factorials at a finite offset — Problem 362's cubic analogue extends this same idea one degree further.