# A limit that tends infinity...

• Jan 16th 2012, 06:11 AM
Flamuri
A limit that tends infinity...
$\displaystyle \lim_{n->\infty } (1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{n^{2}})$

Can someone please help me out here? Have never find the limit of a function like this. I've learned limits but not this kind, if someone could tell me the first steps in those kind of tasks, would be great.

Thanks,
Flamuri
• Jan 16th 2012, 06:16 AM
piscoau
Re: A limit that tends infinity...
well, the comment before is wrong....use trig function to find the limit
• Jan 16th 2012, 06:31 AM
chisigma
Re: A limit that tends infinity...
Quote:

Originally Posted by Flamuri
$\displaystyle \lim_{n->\infty } (1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{n^{2}})$

Can someone please help me out here? Have never find the limit of a function like this. I've learned limits but not this kind, if someone could tell me the first steps in those kind of tasks, would be great.

Thanks,
Flamuri

Remembering the 'infinite product'...

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

... You obtain first...

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

... and from (2)...

$\displaystyle \prod_{n=2}^{\infty} (1-\frac{1}{n^{2}})= \lim_{x \rightarrow 1} \frac{\sin \pi x}{\pi x (1-x^2)}= \frac{1}{2}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$