Limit $\lim_{n\to+\infty} (1-\frac1{2^2})(1-\frac1{3^2})\cdot \cdots \cdot(1-\frac{1}{n^2})$ and series $\sum_{n=2}^{\infty} \ln(1-\frac1{n^2})$ 
Possible Duplicate:
Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$ 

Compute:
\begin{align*}
\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})
\end{align*}
Well, I do so:
$$\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})=\lim_{n\to+\infty}\prod_{j=2}^n (1-\frac{1}{j^2})$$
let $$a_n = \prod_{j=2}^n (1-\frac{1}{j^2})\quad\Rightarrow\quad \ln a_n = \ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)$$
so:
$$\ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)=\sum_{j=2}^{\infty} \ln (1-\frac{1}{j^2})$$
consider
$$\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^2})$$
but as you study this series??

If you write it like
$$
\lim_{n\to \infty}\prod_{j=2}^n \frac{(j+1)(j-1)}{j^2}
$$
you get
$$
\lim_{n\to \infty}\frac{1}{2}\frac{(n+1)!(n-1)!}{(n!)^2},
$$
where factorials cancel in the limit and give $\frac{1}{2}$, because a $2$ is missing in the numerator to get $(n+1)!$.
