"Evaluate the infinite product limn→∞∏k=2n(1−1k2)" solved and explained

Rudy Koch

Answered question

2022-01-24

Evaluate the infinite product

lim_{n \to \infty} \prod_{k = 2}^{n} (1 - \frac{1}{k^{2}})

becky4208fj

Beginner2022-01-25Added 10 answers

Note that

1 - \frac{1}{k^{2}} = (1 - \frac{1}{k}) (1 + \frac{1}{k}) = \frac{k - 1}{k} \frac{k + 1}{k} = \frac{a_{k}}{a_{k - 1}}

with

a_{k} = \frac{k + 1}{k}

, hence this is a telescoping product, i.e.

\prod_{k = 2}^{n} (1 - \frac{1}{k^{2}}) = \frac{a_{2}}{a_{1}} \frac{a_{3}}{a_{2}} \dots \frac{a_{n}}{a_{n - 1}} = \frac{a_{n}}{a_{1}} = \frac{n + 1}{2 n}

ocretz56

Beginner2022-01-26Added 16 answers

Let

g (k) = \frac{k - 1}{k}

. Then this product is:

\prod_{k = 2}^{n} \frac{g (k)}{g (k + 1)}

which is a telescoping product, and thus equal to

\frac{g (2)}{g (n + 1)}

RizerMix

Expert2022-01-27Added 656 answers

(1 - \frac{1}{(k - 1)^{2}}) (1 - \frac{1}{(k)^{2}}) (1 - \frac{1}{(k + 1)^{2}})

= \frac{k (k - 2) (k - 1) (k + 1) k (k + 2)}{(k - 1)^{2} k^{2} (k + 1)^{2}} = \frac{(k - 2) (k + 2)}{(k - 1) (k + 1)}

Observe that

(1 - \frac{1}{(k)^{2}})

is cancelled out by the previous & the next term except for the extreme terms, the 1st & the last term leaving behind the 1st part of the 1st term

= \frac{2 - 1}{2}

and the 2nd part of the last term

= \frac{n + 1}{n}

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