TITLE: $\lim_{n\rightarrow\infty}\ \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\ldots\left(1-\frac{1}{n^2}\right), n\geqslant2 $
QUESTION [6 upvotes]: Calculate the limit of the sequence
$$\lim_{n\rightarrow\infty}\ a_n$$
$$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\ldots\left(1-\frac{1}{n^2}\right), n\geqslant2 $$
Here is what I did:
$\left(1-\frac{1}{n^2}\right)=\left[\left(1-\frac{1}{n^2}\right)^{n^2}\right]^\frac{1}{n^2}=e^\frac{1}{n^2}$
$a_n=e^{\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}}$
$\Rightarrow \lim_{n\rightarrow\infty}\ a_n=e^{\frac{1}{\infty}}=e^0=1$
Not sure if I'm on the right track..
Thank you very much in advance! Really great to see what an amazingly smart community this website has.

REPLY [7 votes]: Hint: We have
$$1-\frac{1}{k^2}=\frac{k^2-1}{k^2}=\frac{k-1}{k}\frac{k+1}{k}.\tag{$1$} $$
Write out the product of the first few terms, in the expanded form described in $(1)$.  There will be a whole lot of collapsing (cancellation) going on.