# How to solve this : $\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$

How to find the sum of this :

$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$$

My Working :

$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1} = 1-0 = 1$$

Is it correct

-
It would have been true if $\prod_{n=1}^{\infty} (a_n-b_n) = \prod_{n=1}^{\infty} a_n - \prod_{n=1}^{\infty} b_n$. Sadly, it's not... – user66258 Aug 7 '13 at 17:14
Try factorization of numerator and denominator. – Kunnysan Aug 7 '13 at 17:15
What @Kunnysan, plus note that $(n+1)^2-(n+1)+1 = n^2+n+1$. – Thomas Andrews Aug 7 '13 at 17:21
Maple produces $$product((1-1/n^3)/(1+1/n^3), n = 2 .. infinity)$$ $$\frac 2 3 ,$$ product(1-1/n^3, n = 2 .. infinity) $$1/3\,{\frac {\sin \left( \pi \, \left( 1/2+1/2\,i\sqrt {3} \right) \right) }{\pi }},$$ and $$product(1+1/n^3, n = 2 .. infinity)$$ $$1/2\,{\frac {\sin \left( \pi \, \left( 1/2+1/2\,i\sqrt {3} \right) \right) }{\pi }}.$$ The imaginary parts equal zero. – user64494 Aug 7 '13 at 17:28
From where can I get the details on telescoping series.. please suggest thanks... to all of you.. – sultan Aug 7 '13 at 17:29

HINT:

If $$t_n=\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$

$$t_{n+1}=\frac{n\{(n+1)^2+n+1+1\}}{(n+2)\{(n+1)^2-(n+1)-1\}}=\frac{n\{(n+1)^2+n+1+1\}}{(n+2)(n^2+n+1)}$$

and $$t_{n-1}=\frac{(n-2)\{(n-1)^2+n-1+1\}}{n\{(n-1)^2-(n-1)-1\}}=\frac{(n-2)(n^2-n+1)}{n\{(n-1)^2-(n-1)-1\}}$$

Alternatively, let $\displaystyle u_n=\frac{n-1}{n+1},v_n=\frac{n^2+n+1}{n^2-n+1}$ so that $t_n=u_n\cdot v_n$

$$\implies\prod_{2\le n\le r}u_n=\frac{1\cdot2\cdot3\cdots(r-2)(r-1)}{3\cdot4\cdot5\cdots r(r+1)}\frac2{r(r+1)}$$

$$\implies\prod_{2\le n\le r}v_n=\frac{7\cdot13\cdot21\cdots(r^2-r+1)(r^2+r+1)}{3\cdot7\cdot13\cdots \{(r-1)^2-(r-1)+1\}(r^2-r+1)}=\frac{r^2+r+1}3$$

$$\implies\prod_{2\le n\le r} t_n=\left(\prod_{2\le n\le r}u_n\right)\left( \prod_{2\le n\le r}v_n\right)=\frac{2(r^2+r+1)}{3r(r+1)}$$

Setting $r\to\infty,$ $\displaystyle \frac{2(r^2+r+1)}{3r(r+1)}=\frac23\lim_{r\to\infty}\frac{1+\frac1r+\frac1{r^2}}{1+\frac1r}=?$

-
From where can I get the details on telescoping series.. please suggest thanks... to all of you.. – sultan Aug 7 '13 at 17:30
You mean, other than googling "Telescoping Series" and finding the Wikipedia page? en.wikipedia.org/wiki/Telescoping_series – Thomas Andrews Aug 7 '13 at 17:32
@sultan, just set $n=1,2,3,\cdots$ in the first method to see which terms survive. My hint shows that there will be ample cancellations. Also, please find the alternative method – lab bhattacharjee Aug 7 '13 at 18:56

Let $f(n) =\dfrac{n(n-1)}{n^2-n+1}$. Then show $f(n+1) = \dfrac{n(n+1)}{n^2+n+1}$ and thus $$\frac{f(n)}{f(n+1)} = \frac{n(n-1)(n^2+n+1)}{n(n+1)(n^2-n+1)} = \frac{n^3-1}{n^3+1}$$
(I call this a "magic answer" just because most of the other answers here give you reasons for how you would see this, while I just plop in an $f$ that works, as if by magic. This is really the same argument as the others, just distilled to a minimalist essence.)
Note that $${n^3 - 1 \over n^3 + 1} = {n - 1 \over n + 1}{n^2 + n + 1 \over n^2 - n + 1}$$ Also note that $$(n-1)^2 + (n - 1) + 1 = n^2 - n + 1$$ So the infinite product in question is really the product of two telescoping products.