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How do I solve the infinite product of $$\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}?$$

I know that I have to factorise to $$\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)},$$

but how do I do the partial product?

Thanks a lot in advance.

If I'm not mistaken the Answer is 2/3

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Thanks for the edit @Cameron Buie – Ray Jul 11 '12 at 18:58
Calculus class - – Ray Jul 11 '12 at 19:05
As frequently happens, the word "solve" is getting used here as a catch-all term when it's not correct. One solves equations; one solves problems. One evaluates or finds expressions. – Michael Hardy Jul 11 '12 at 19:42
up vote 7 down vote accepted

After factorization, the product looks like $(\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}\frac{5}{7}\cdots)(\frac{7}{3}\frac{13}{7}\frac{21}{13}\frac{31}{21}\frac{43}{31}\cdots)=(2)(1/3)=2/3$.$$$$ Here terms in first () are from expression $\frac{n-1}{n+1}$ and terms in second () from expression $\frac{n^2+n+1}{n^2-n+1}$ .

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Thanks but why the first () becomes 2 and the 2nd () becomes 1/3? – Ray Jul 11 '12 at 19:06
As you can see in first () all terms other than $1$ and $2$ comes in both numerator and denominator and thus cancels out leaving only $1\times 2$. Similarly, in second () all terms other than $3$ in denominator cancels out leaving product as $1/3.$ – Aang Jul 11 '12 at 19:09
Thank you very much! – Ray Jul 11 '12 at 19:10

Hint: Let $f(x)=x^2+x+1$. Then $f(n)=n^2+n+1$, and $f(n-1)=n^2-n+1$. This will enable you to "telescope" the terms $\frac{n^2+n+1}{n^2-n+1}$. A whole lot of cancelling going on.

The $\frac{n-1}{n+1}$ terms also telescope.

I would suggest that you write down the terms you are multiplying, for $n=2$, $3$, $4$, even $5$. Express each term in the factored form mentioned in the post. For example, for $n=2$ we will have $\frac{1\cdot 7}{3\cdot 3}$. The collapse will be visually clear.

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Thanks a lot for explaining that! – Ray Jul 11 '12 at 19:09

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