# Find the product of the series? [closed]

$$\prod_{n=2}^\infty\dfrac{n^3 - 1}{n^3 + 1}$$

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## closed as off-topic by Sami Ben Romdhane, TooTone, Claude Leibovici, Davide Giraudo, vonbrandMar 4 '14 at 17:24

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Calculate the first few partial products. You should spot a pattern not too far in. – Daniel Fischer Mar 4 '14 at 15:40
...please...? And what have you tried? – DonAntonio Mar 4 '14 at 15:43
This is a duplicate of this question. – robjohn Mar 8 '14 at 0:42

Notice $$\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$ and $(n+1)^2-(n+1)+1=n^2+n+1$. Therefore we have a telescoping product
$$\prod_{n=2}^\infty \frac{n^3-1}{n^3+1}=\frac{1\cdot \color{blue}{7}}{\color{red}{3}\cdot 3}\cdot\frac{2\cdot \color{blue}{13}}{\color{red}{4}\cdot \color{blue}{7}}\cdot \frac{\color{red}{3}\cdot \color{blue}{21}}{\color{red}{5}\cdot \color{blue}{13}}\cdots=\frac{1\cdot 2}{3}=\frac{2}{3}$$
Note that $$\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}=\frac{(n-1)}{(n+1)}\cdot\frac{\big((n+1)^2-(n+1)+1\big)}{(n^2-n+1)}.$$ We have $$\prod_{n=2}^N\frac{n-1}{n+1}=\frac{1\cdot 2\cdots(N-2)(N-1)}{3\cdot 4\cdots N(N+1)}=\frac{1\cdot 2}{N(N+1)},$$ while \begin{align} \prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}&=\prod_{n=2}^N\frac{\big((n+1)^2-(n+1)+1\big)}{(n^2-n+1)}\\&=\frac{\big((N+1)^2-(N+1)+1\big)}{2^2-2+1}=\frac{N^2+N+1}{3}, \end{align} and hence $$\prod_{n=2}^N\frac{n^3-1}{n^3+1}=\frac{1\cdot 2\cdot(N^2+N+1)}{3\cdot N(N+1)}\to \frac{2}{3},$$ as $N\to \infty$.