Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
6  
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

3 Answers 3

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

share|improve this answer
    
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

Magic answer:

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.)

share|improve this answer

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.

share|improve this answer
    
From where can I get the details on telescoping series.. please suggest thanks... to all of you.. –  sultan Aug 7 '13 at 17:23
    
The idea is that if you multiply the terms of a telescoping series, the numerator of a given term will be cancelled by a corresponding denominator from a different term. So multiplying them together will lead to a lot of cancellation, and you will be able to find the limit quite easily. –  Zarrax Aug 7 '13 at 17:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.