Find limit of this decreasing sequence $$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots 
\left(1-\frac{1}{n^2}\right)
$$
I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit. 
 A: Answer:
$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$
After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$
$$\frac{n+1}{2}\cdot\frac{1}{n}$$
Limit of this function tending to infinity $= 1/2$.
A: Hint: try defining $b_n = \ln(a_n)$ (which is well-defined) and see what limit this goes to. Then use a certain exponential function to see what $\lim_n a_n$ is.
A: Hint: Rewrite each $1-\frac{1}{k^2}$ as $\frac{(k-1)(k+1)}{k^2}$ and observe the mass cancellations. It will be useful to do this explicitly for say the product of the first $5$ terms. 
A: We write
$$a_n=\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{(k-1)(k+1)}{k^2}=\prod_{k=2}^n\frac{v_k}{v_{k+1}}$$
where
$$v_k=\frac{k-1}{k}$$
so by change of index
$$a_n=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=2}^nv_{k+1}}=\frac{\displaystyle\prod_{k=2}^nv_k}{\displaystyle\prod_{k=3}^{n+1}v_{k}}=\frac{v_2}{v_{n+1}}=v_2\times\frac{n+1}{n}\to v_2=\frac12$$
