The convergence of an infinite seqeunce Suppose that 
$$
a_n = \prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right),
$$ 
for $n \geq 2$. How can we show that 
$$
\lim_{n \to \infty} a_n = \lim_{n \to \infty}\prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right) = 1?
$$ 
Thanks very much.
 A: prove that $\prod_{k=n}^{m}1-\frac{1}{k^2}=\frac{(1+m)(n+1)}{mn}$
A: Consider the following.
Method 1

The product is
\begin{align}
\prod_{k=n}^{\infty} \left(1 - \frac{1}{k^2}\right) = \frac{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{n-1} \left(1 - \frac{1}{k^2}\right) }
\end{align}
for which
\begin{align}
\lim_{n \rightarrow \infty} \, \prod_{k=n}^{\infty}\left(1 - \frac{1}{k^2}\right) &= \lim_{n \rightarrow \infty} \frac{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{n-1} \left(1 - \frac{1}{k^2}\right) } \\
&= \frac{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{\infty}\left(1 - \frac{1}{k^2}\right) } = 1 
\end{align}
Method 2

It can be determined that
\begin{align}
\prod_{k=2}^{m} \left(1 - \frac{1}{k}\right) &= \frac{1}{m} \\
\prod_{k=2}^{m}\left(1 + \frac{1}{k}\right) &= \frac{m+1}{2} 
\end{align}
such that 
\begin{align}
\prod_{k=2}^{m} \left( 1 - \frac{1}{k^{2}} \right) = \frac{m+1}{2 m}.
\end{align}
Now,
\begin{align}
\lim_{n \rightarrow \infty} \prod_{k=n}^{\infty} \left(1 - \frac{1}{k^2}\right) &= \lim_{m,n \rightarrow \infty} \prod_{k=n}^{2m} \left(1 - \frac{1}{k^2}\right) \\
&= \lim_{m,n \rightarrow \infty} \frac{ \prod_{k=2}^{2m} \left(1 - \frac{1}{k^2}\right) }{ \prod_{k=2}^{n-1} \left(1 - \frac{1}{k^2}\right) } \\
&= \lim_{m,n \rightarrow \infty} \, \frac{ \frac{2m+1}{4m} }{ \frac{n}{2(n-1)} } \\
&= \lim_{m,n \rightarrow \infty} \left( 1 + \frac{1}{2m} \right) \left( 1 - \frac{1}{n} \right) = 1.   
\end{align}
