A lusty product

Using $\small{\color{#3D99F6}{a^3+b^3=(a+b)(a^2-ab+b^2)}}$ $P= \displaystyle \prod_{n=2}^{m} \dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$ $=(\displaystyle \prod_{n=2}^{m} \dfrac{n-1}{n+1})\times (\displaystyle \prod_{n=2}^{m} \dfrac{n^2+n+1}{n^2-n+1})$ $=(\dfrac{1}{3}\times \dfrac{2}{4}\times \dfrac{3}{5}\times \dfrac{4}{6}\times....\dfrac{m-1}{m+1})\times (\dfrac{7}{3}\times \dfrac{13}{7}\times \dfrac{21}{13}\times \dfrac{31}{21}\times...\dfrac{m^2+m+1}{m^2-m+1})$ $=\dfrac{2(m^2+m+1)}{3(m)(m+1)}$ When m $\rightarrow \infty,\color{forestgreen}{P~\text{approaches}~\dfrac{2}{3}}$

Hence, $\Large\color{goldenrod}{2+3=5}$

Let's consider the general term $a_n \equiv \frac{n^3-1}{n^3+1} = \frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} = \frac{(n-1)+n(n-1)(n+1)}{(n+1)+n(n-1)(n+1)} = \frac{1 + \frac{1}{n(n+1)}}{1 + \frac{1}{n(n-1)}}\;.$ Let's call: $x_n \equiv 1 + \frac{1}{n(n - 1)}\;.$ We can write the original product as $\prod_{n=2}^{\infty}a_n = \prod_{n=2}^{\infty}\frac{x_{n+1}}{x_n}\;.$ Numerators and denominators of consecutive terms therefore simplify. The partial product $P_N \equiv \prod_{n=2}^N\frac{x_{n+1}}{x_n} = \frac{x_{N+1}}{x_2} = \frac{1 + \frac{1}{N(N + 1)}}{1 + \frac{1}{2(2 - 1)}} = \frac{2}{3}\left[1 + \frac{1}{N(N + 1)}\right]\;.$ Thus $\lim_{N\to\infty}P_N = \frac{2}{3}\;.$

Consider:

$\begin{aligned} \prod_{n=2}^\infty \frac{n^3-1}{n^3+1} & = \prod_{n=2}^m \frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ & = \small \frac{1 \dot{} \color{#3D99F6}{\not{7}}}{\color{#D61F06}{\not{3}} \dot{}3} \dot{} \frac{2 \dot{} \color{#3D99F6}{\not{13}}}{\color{#D61F06}{\not{4}} \dot{} \color{#3D99F6}{\not{7}}} \dot{} \frac{\color{#D61F06}{\not{3}} \dot{} \color{#3D99F6}{\not{21}} }{\color{#D61F06}{\not{5}} \dot{} \color{#3D99F6}{\not{13}}} \dot{} \frac{\color{#D61F06}{\not{4}} \dot{} \color{#3D99F6}{\not{31}} }{\color{#D61F06}{\not{6}} \dot{} \color{#3D99F6}{\not{21}}} \dot{} \frac{\color{#D61F06}{\not{5}} \dot{} \color{#3D99F6}{\not{43}} }{\color{#D61F06}{\not{7}} \dot{} \color{#3D99F6}{\not{31}}} \dot{} ... \\ & = \frac{2}{3} \end{aligned}$

$\Rightarrow a + b = 2 + 3 = \boxed{5}$

nice problem.

An old classic Putnam problem......Question A1 from the '77 Exam!