Prove that product from n=2 to infinity
Prove that $\displaystyle\prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}=\frac23$.
Explanation of Method
Factoring via sum/difference of cubes, $n^3-1=(n-1)(n^2+n+1)$ and $n^3+1=(n+1)(n^2-n+1)$, splits the product into a simple telescoping linear part and a quadratic part that telescopes once it's noticed that $n^2+n+1=f(n+1)$ for $f(m)=m^2-m+1$.
- Factor using $a^3\mp1=(a\mp1)(a^2\pm a+1)$.
- Telescope the linear part $\frac{n-1}{n+1}$ directly.
- Telescope the quadratic part by recognizing $n^2+n+1=f(n+1)$, then combine and take the limit.
Step 1 · Factor
$$ n^3-1=(n-1)(n^2+n+1),\qquad n^3+1=(n+1)(n^2-n+1), $$ $$ P_N=\prod_{n=2}^N\frac{n^3-1}{n^3+1}=\left(\prod_{n=2}^N\frac{n-1}{n+1}\right)\left(\prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}\right). $$
Step 2 · The linear telescoping product
$$ \prod_{n=2}^N\frac{n-1}{n+1}=\frac13\cdot\frac24\cdot\frac35\cdots\frac{N-1}{N+1}=\frac{1\cdot2}{N(N+1)}=\frac2{N(N+1)}. $$
Step 3 · The quadratic telescoping product
With $f(m)=m^2-m+1$, note $n^2+n+1=f(n+1)$, so $$ \prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}=\prod_{n=2}^N\frac{f(n+1)}{f(n)}=\frac{f(N+1)}{f(2)}=\frac{N^2+N+1}3. $$ Combining: $$ P_N=\frac2{N(N+1)}\cdot\frac{N^2+N+1}3=\frac{2(N^2+N+1)}{3N(N+1)}. $$ $$ \boxed{\lim_{N\to\infty}P_N=\lim_{N\to\infty}\frac{2\bigl(1+1/N+1/N^2\bigr)}{3\bigl(1+1/N\bigr)}=\frac23} $$ (Numerical check: $P_2=7/9\approx0.778$, $P_{10}=37/55\approx0.673$, $P_{100}\approx0.667$; direct product to $10^5$ terms gives $0.666667$.)
Summary Table
| Property | Value |
|---|---|
| Partial product | $P_N=\dfrac{2(N^2+N+1)}{3N(N+1)}$ |
| Limit | $\dfrac23$ |
Additional Notes
- This is a cubic-degree analogue of Problem 361's quadratic telescoping: both use the identity $a^k\mp1=(a\mp1)(\ldots)$ to split into a simple linear telescoping factor plus a higher-degree telescoping factor.
- The key trick — recognizing $n^2+n+1$ as a shift $f(n+1)$ of $n^2-n+1=f(n)$ — is the cubic-product analogue of the shifted-index telescoping seen in Problems 360 and 361, and generalizes to any product of the form $\prod\frac{g(n+1)}{g(n)}$ for polynomial $g$.
- Since $\frac{n^3-1}{n^3+1}\to1$ like $1-2/n^3$ for large $n$, and $\sum1/n^3$ converges (a $p$-series with $p=3>1$), this product converges absolutely to a nonzero limit — consistent with the general convergence criterion applied in Problem 370.