Thread: A limit that tends infinity...

1. A limit that tends infinity...

$\lim_{n->\infty } (1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{n^{2}})$

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

2. Re: A limit that tends infinity...

well, the comment before is wrong....use trig function to find the limit

3. Re: A limit that tends infinity...

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

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'...

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

... You obtain first...

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

... and from (2)...

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

Kind regards

$\chi$ $\sigma$