Telescoping Cubes

Calculus Level 3

$\large \displaystyle \prod_{n=2}^{\infty} \dfrac {n^3-1}{n^3 + 1}$

If the value of the above expression can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

The answer is 5.

2 solutions

Chew-Seong Cheong
Apr 4, 2016

$\begin{aligned} P & = \prod_{n=2}^\infty \frac{n^3-1}{n^3+1} \\ & = \prod_{n=2}^\infty \frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ & = \frac{\displaystyle \color{#3D99F6}{\prod_{n=2} ^\infty (n-1)} \prod_{n=2} ^\infty \left[ \left(n + \frac{1}{2}\right)^2 + \frac{3}{4}\right]}{\displaystyle \prod_{n=2} ^\infty (n+1) \color{#D61F06}{\prod_{n=2} ^\infty \left[\left(n - \frac{1}{2}\right)^2 + \frac{3}{4}\right]}} \\ & = \frac{\displaystyle \color{#3D99F6}{1\cdot{} 2 \cdot{} \prod_{n=2} ^\infty (n+1)} \cdot{} \prod_{n=2} ^\infty \left[ \left(n + \frac{1}{2}\right)^2 + \frac{3}{4}\right]}{\displaystyle \prod_{n=2} ^\infty (n+1) \cdot{} \color{#D61F06}{ \left[\left(2 - \frac{1}{2}\right)^2 + \frac{3}{4}\right] \prod_{n=2} ^\infty \left[\left(n + \frac{1}{2}\right)^2 + \frac{3}{4}\right]}} \\ & = \frac{1 \cdot{} 2}{\left(\frac{3}{2}\right)^2 + \frac{3}{4}} = \frac{2}{\frac{9}{4}+\frac{3}{4}} = \frac{2}{3} \end{aligned}$

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

Sayandeep Ghosh
Apr 4, 2016

Same problem given by Aditya sharma

This problem has been posted many a times.

Rohit Udaiwal - 5 years, 2 months ago

Exactly..... Many a times this problem has been posted

Rishabh Jain - 5 years, 2 months ago

Telescoping Cubes

The answer is 5.

2 solutions

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