How to calculate the limit of this product?

**gotmejerry** · December 18th 2011, 06:41 AM

$\prod_{n=2}^{\infty}(1-\frac{1}{n^2})=\lim_{n\to\infty}(1-\frac{1}{4})(1-\frac{1}{9})\dots (1-\frac{1}{n^2})$

Thank you!

**ILikeSerena** · December 18th 2011, 07:10 AM

Hi gotmejerry!

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

Can you write out a couple of factors and see how things cancel?

**gotmejerry** · December 18th 2011, 07:38 AM

No unfortunately I cant.

**ILikeSerena** · December 18th 2011, 07:39 AM

Which factor do you get if you fill in n=2?
And which factor for n=3?
And n=4?

**gotmejerry** · December 18th 2011, 07:41 AM

3/4 8/9 15/16 but what do i do with them, how do they cancel each other?

**ILikeSerena** · December 18th 2011, 07:46 AM

Let's write that out:

${1 \cdot 3 \over 2^2} \cdot {2 \cdot 4 \over 3^2} \cdot {3 \cdot 5 \over 4^2} \cdot {4 \cdot 6 \over 5^2} \cdot ... \cdot {(n-1) (n+1) \over n^2}$

Do you see numbers that cancel?

**gotmejerry** · December 18th 2011, 07:54 AM

Yes, thanks

, so If I am not wrong, only 1/2 *(n+1)/n remains? And its limit is 1/2. Thank you very much

**ILikeSerena** · December 18th 2011, 07:56 AM

Right!

**chisigma** · December 18th 2011, 08:14 AM

Originally Posted by gotmejerry

$\prod_{n=2}^{\infty}(1-\frac{1}{n^2})=\lim_{n\to\infty}(1-\frac{1}{4})(1-\frac{1}{9})\dots (1-\frac{1}{n^2})$

Thank you!

Remembering the 'infinite product'...

$\frac{\sin \pi x}{\pi x}= \prod_{n=1}^{\infty} (1- \frac{x^{2}}{n^{2}})$ (1)

... Your 'infite product' seems to be [applying l'Hopital rule] ...

$\prod_{n=2}^{\infty} (1- \frac{1}{n^{2}}) = \lim_{x \rightarrow 1} \frac{\sin \pi x}{\pi x (1-x^{2})} = \frac{1}{2}(2)$

Marry Christmas from Serbia

$\chi$ $\sigma$

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