Can you multiply to infinity?

Calculus Level 3

Evaluate the given expression for Q:

$\displaystyle\prod _{ n=2 }^{ \infty }{ (1-\frac { 1 }{ { n }^{ 2 } } ) }$ = $\textbf{Q}$

If $\textbf{Q}$ can be represented as $\frac { a }{ b }$ , where $\textbf{a}$ and $\textbf{b}$ are coprime integers, then what is $a+b$ ?

The answer is 3.

2 solutions

Mehul Chaturvedi
Jan 27, 2015

The product of the first $N - 1$ terms can be written as $\displaystyle\prod_{n=2}^{N} \dfrac{n^{2} - 1}{n^{2}} = \prod_{n=2}^{N} \dfrac{(n - 1)(n + 1)}{n*n}$ .

Now we'll get a series like this

$\dfrac{(1)\times(3)}{2\times 2}\times\dfrac{(2)\times(4)}{3\times 3}\times\dfrac{(3)\times(4)}{4\times 4}........$ and so on

Now all the 'middle' terms cancel, leaving us with the product

$\dfrac{1}{2} * \dfrac{N+1}{N}$ , which as $N \rightarrow \infty$ goes to $\dfrac{1}{2}$ . Thus $a + b = 1 + 2 = \boxed{\huge\color{royalblue}{{3}}}$ .

Again, a formal way to represent this is to demonstrate the index shifting in the product notation as:

$Q=\prod_{n=2}^\infty \dfrac{(n-1)(n+1)}{n^2} =\left(\prod_{n=2}^\infty \dfrac{(n-1)}{n}\right)\left(\prod_{n=3}^\infty \dfrac{n}{(n-1)}\right)$

$\implies Q=\dfrac{1}{2}\cdot \prod_{n=3}^\infty \left(\dfrac{(n-1)}{n}\cdot \dfrac{n}{(n-1)}\right) = \dfrac{1}{2}\cdot \prod_{n=3}^\infty\left(1\right) = \dfrac{1}{2}$

Prasun Biswas - 6 years, 4 months ago

Brian Charlesworth
Jan 5, 2015

The product of the first $N - 1$ terms can be written as

$\displaystyle\prod_{n=2}^{N} \dfrac{n^{2} - 1}{n^{2}} = \prod_{n=2}^{N} \dfrac{(n - 1)(n + 1)}{n\times n}$ .

Written in this form it is clear that we have a telescoping product, where all the 'middle' terms cancel, leaving us with the product

$\dfrac{1}{2} * \dfrac{N+1}{N}$ , which as $N \rightarrow \infty$ goes to $\dfrac{1}{2}$ .

Thus $a + b = 1 + 2 = \boxed{3}$ .

Can you multiply to infinity?

The answer is 3.

2 solutions

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