6
$\begingroup$

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

  • 7
    As frequently happens, the word "sol$v$e" 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.2012-07-11

2 Answers 2

7

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 $\frac{n^2+n+1}{n^2-n+1}$ .

  • 0
    One should note here that $(\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}\frac{5}{7}\cdots)\to 0$ while $(\frac{7}{3}\frac{13}{7}\frac{21}{13}\frac{31}{21}\frac{43}{31}\cdots)\to\infty$.2017-02-09
9

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.

  • 0
    Thanks a lot for explaining that!2012-07-11