## Tuesday, May 26, 2015

### Putnam Math Exam Problem: Evaluate [MATH]\prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}[/MATH]

Putnam Math Exam Problem:

Evaluate [MATH]\prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}[/MATH].

I admit it, the first (or should I say, the only) thing that we could do about the given fraction's expression, is to factor it:

$\dfrac{n^3-1}{n^3+1}=\dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$

That's it!

In order to proceed with a good plan, we need to at least suspect a thing. We need to suspect that the expressions on the numerator and denominator are related as a telescoping product.

If things aren't immediately obvious to us, what we could do is to plug in $n=2,\,3,\,4,\,\cdots,\,m$ into the fraction and see where it leads us:

[MATH]\prod_{n=2}^{m}\frac{n^3-1}{n^3+1}[/MATH]

[MATH]=\prod_{n=2}^{m}\dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}[/MATH]

[MATH]\scriptsize=\left(\dfrac{(2-1)(2^2+2+1)}{(2+1)(2^2-2+1)}\right)\left(\dfrac{(3-1)(3^2+3+1)}{(3+1)(3^2-3+1)}\right)\left(\dfrac{(4-1)(4^2+4+1)}{(4+1)(4^2-4+1)}\right)\cdots\left(\dfrac{((m-1)-1)((m-1)^2+(m-1)+1)}{((m-1)+1)((m-1)^2-(m-1)+1)}\right)\left(\dfrac{(m-1)(m^2+m+1)}{(m+1)(m^2-m+1)}\right)[/MATH]

[MATH]=\left(\dfrac{(1)(7)}{(3)(3)}\right)\left(\dfrac{(2)(13)}{(4)(7)}\right)\left(\dfrac{(3)(21)}{(5)(13)}\right)\cdots\left(\dfrac{((m-2)(m^2-m+1)}{(m)(m^2-3m+3)}\right)\left(\dfrac{(m-1)(m^2+m+1)}{(m+1)(m^2-m+1)}\right)[/MATH]

[MATH]\scriptsize=\left(\dfrac{(1)\color{blue}\cancel{(7)}}{\color{red}\cancel{(3)}\color{black}(3)}\right)\left(\dfrac{(2)\color{blue}\cancel{(13)}}{\color{red}\cancel{(4)}\color{blue}\cancel{(7)}}\right)\left(\dfrac{\color{red}\cancel{(3)}\color{blue}\cancel{(21)}}{\color{red}\cancel{(5)}\color{blue}\cancel{(13)}}\right)\cdots\left(\dfrac{\color{red}\cancel{(m-2)}\color{blue}\cancel{(m^2-m+1)}}{(m)\color{blue}\cancel{(m^2-3m+3)}}\right)\left(\dfrac{\color{red}\cancel{(m-1)}\color{black}(m^2+m+1)}{(m+1)\color{blue}\cancel{(m^2-m+1)}}\right)[/MATH]

$=\dfrac{2(m^2+m+1)}{3(m)(m+1)}$

$=\dfrac{2(m^2+m+1)}{3(m^2+m)}$

$=\dfrac{2(m^2+m+1)\div m^2}{3(m^2+m)\div m^2}$

$=\dfrac{2\left(1+\dfrac{1}{m}+\dfrac{1}{m^2}\right)}{3\left(1+\dfrac{1}{m}\right)}$

Now, taking the limit as $m \to \infty$, we get:

[MATH]\begin{align*}\prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}&=\lim_{{m}\to{\infty}}\dfrac{2\left(1+\dfrac{1}{m}+\dfrac{1}{m^2}\right)}{3\left(1+\dfrac{1}{m}\right)}\\&=\dfrac{2\left(1+0+0\right)}{3\left(1+0\right)}\\&=\dfrac{2}{3}\end{align*}[/MATH]

In other words, we need to at least get started to work on the problem by doing what we're afford to do, and in this instance, that simply means we need to factor the expressions on the numerator and denominator.

More often than not, we would end up hitting on good ideas upon finished doing what our intuition asked us to do in the first step.