Putnam Product!

$n^3 - 1 = \left(n-1\right) \left(n^2+n+1\right)$

$n^3 + 1 = \left(n+1\right) \left(n^2-n+1\right)$

From this we see that $\prod_{n=2}^\infty \frac{n^3 -1}{n^3 +1}= \prod_{n=2}^\infty \frac{n-1}{n+1}\prod_{n=2}^\infty \frac{n^2+n+1}{n^2-n+1}$

Let: $f \left(n \right) = n-1, \: g \left(n\right) = n+1, \: h \left(n\right) = n^2 + n + 1, \: i\left(n\right) = n^2 -n +1$

Clearly $f\left(n+2\right) = g\left(n\right)$ therefore $\prod_{n=2}^\infty \frac{n-1}{n+1} = \frac{1}{3}*\frac{2}{4}*\frac{3}{5}*\frac{4}{6}...=2$

We also see that

$i\left(n+1\right) = \left(n+1\right)^2 - \left(n+1\right) + 1$

$= n^2 + 2n + 1 -n - 1 + 1$

$= n^2 + n +1$

$= h\left(n\right)$

Therefore $\prod_{n=2}^\infty \frac{n^2+n+1}{n^2-n+1} = \frac{7}{3}*\frac{13}{7}*\frac{21}{13}... = \frac{1}{3}$

Hence we get $\prod_{n=2}^\infty \frac{n^3-1}{n^3+1} = \frac{2}{3}$

So we get $p+q = 2+3 = 5$

$\begin{aligned} \prod_{n=2}^\infty \dfrac{n^3-1}{n^3+1} &=&\lim_{N\to \infty} \left[\prod_{n=2}^N \dfrac{n^3-1}{n^3+1}\right] \\ &=&\lim_{N\to \infty} \left[\prod_{n=2}^N \dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\right] \\ &=& \lim_{N\to \infty} \left[\prod_{n=2}^N \dfrac{n-1}{n+1}\prod_{n=2}^N \dfrac{n^2+n+1}{n^2-n+1}\right] \\ &=& \lim_{N\to \infty} \left[\dfrac{2(N-1)!}{(N+1)!}\prod_{n=2}^N \dfrac{n^2+n+1}{n^2-n+1}\right] \\ &=& \lim_{N\to \infty} \left[\dfrac{2}{N(N+1)}\prod_{n=2}^N \dfrac{n^2+n+1}{n^2-n+1}\right] \end{aligned}$

Since: $n^2+n+1=(n+1)^2-(n+1)+1$ , the latter product telescopes too (all the terms except the first index term and the last index term).

$\begin{aligned} \lim_{N\to \infty} \left[\dfrac{2}{N(N+1)}\prod_{n=2}^N \dfrac{n^2+1+n}{n^2+1-n}\right] &=& \lim_{N\to \infty} \left[\dfrac{2}{N(N+1)}\times\dfrac{N^2+N+1}{3}\right] \\ &=& \lim_{N\to\infty} \left[\dfrac{2}{3} \times \dfrac{N^2+N+1}{N^2+N}\right] \\ &=& \lim_{N\to\infty} \left[\dfrac{2}{3} \left(1+\dfrac{1}{N^2+N}\right)\right] \\ &=& \boxed{\dfrac{2}{3}} \equiv \dfrac{p}{q}\\ \end{aligned}$

$\therefore ~~~ p+q=2+3=\fbox{5}$

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

= $\frac { { 2 }^{ 3 }\quad -\quad 1 }{ { 2 }^{ 3 }\quad +\quad 1 } \times \frac { { 3 }^{ 3 }\quad -\quad 1 }{ { 3 }^{ 3 }\quad +\quad 1 } \times ........\infty$ [ $\prod _{ }^{ }{ }$ means product of each term]

Using formula a^3 - b^3 = (a-b)(a^2 + b^2 +ab) and a^3 + b^3 = (a+b)(a^2 + b^2 -ab),

= $\frac { (2-1)({ 2 }^{ 2 }\quad +\quad { 1 }^{ 2 }\quad +2*1) }{ (2+1)({ 2 }^{ 2 }\quad +\quad { 1 }^{ 2 }\quad -2*1) } \times \frac { (3-1)({ 3 }^{ 2 }\quad +\quad { 1 }^{ 2 }\quad +3*1) }{ (3+1)({ 3 }^{ 2 }\quad +\quad { 1 }^{ 2 }\quad -3*1) } \times \frac { (4-1)({ 4 }^{ 2 }\quad +\quad { 1 }^{ 2 }\quad +4*1) }{ (4+1)({ 4 }^{ 2 }\quad +\quad { 1 }^{ 2 }\quad -4*1) } \times ....\infty$

On simplifying, we get,

= $\frac { 1\times 7 }{ 3\times 3 } \times \frac { 2\times 13 }{ 4\times 7 } \times \frac { 3\times 21 }{ 5\times 13 } \times ....\infty$

therefore, 7 and 7, 3 and 3, 13 and 13 will cut and this trend will go on forever as it is an infinite series and so, every term will cut except 2 and 3

= $\boxed { \frac { 2 }{ 3 } }$

Hence, the answer is 5