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

Question

The value of the infinite product $$P = \frac 79 \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots$$ is

(A) $1$
(B) $2/3$
(C) $7/3$
(D) none of the above

I wrote first 6 terms and tried to cancel out but did not get any idea what will be the last term

I did one same kind of problem $\prod (1-{1\over k^2}),k\ge2$ whose answer is $1\over 2$

here I think the answer will be $D$, none of this?

Answer 1

Hint: It's easier if you rewrite it as:

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

Then note that $(n-1)^2+(n-1)+1 = n^2-n+1$.

Answer 2

HINT:

Observe that $$n^3-1=(n-1)(n^2+n+1)\text{ and }n^3+1=(n+1)(n^2-n+1)$$

Again, $\displaystyle (n+1)^2-(n+1)+1=\cdots=n^2+n+1$

Just set a few values of $n$ to find the surviving terms

Answer 3

Hint: $$ \begin{align} \prod_{n=2}^\infty\frac{n^3-1}{n^3+1} &=\lim_{m\to\infty}\prod_{n=2}^m\frac{n-1}{n+1}\frac{n^2+n+1}{n^2-n+1}\\ \end{align} $$ The limit of partial products overcomes the difficulty of multiplying two divergent products that an old answer had.

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Telescoping Products

Look for terms that cancel: $$ \prod_{n=2}^m\frac{n-1}{n+1}=\frac13\frac24\frac35\frac46\cdots\frac{m-3}{m-1}\frac{m-2}{m}\frac{m-1}{m+1} $$ $$ \prod_{n=2}^m\frac{n^2+n+1}{n^2-n+1}=\frac73\frac{13}7\frac{21}{13}\frac{31}{21}\cdots\frac{m^2-m+1}{m^2-3m+3}\frac{m^2+m+1}{m^2-m+1} $$