Q:

What is the limit of the sequence $a_n=(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdots(1-\frac{1}{n^2})$?

I am stuck on finding the limit of
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$$
Can anybody help?
UPDATE
Is there a solution involving making the fractions like
$$(\frac{n^2-1}{n^2})$$
and using the formula
$$(2^23^24^2...n^2) = (n!)^2$$?

A:

Since, for every $k$, $$1 - \frac{1}{k^2} = \left(1 - \frac{1}{k}\right)\left(1 + \frac{1}{k}\right) = \frac{k-1}{k}\frac{k+1}{k} = \dfrac{\frac{k+1}{k}}{\frac{k}{k-1}}$$ we have by telescoping $$\left(1 - \frac{1}{2^2}\right)\cdots \left(1 - \frac{1}{n^2}\right) = \prod_{k = 2}^n \dfrac{\frac{k+1}{k}}{\frac{k}{k-1}} = \dfrac{\frac{n+1}{n}}{\frac{2}{2-1}} = \frac{n+1}{2n}.$$ So the limit is $\frac{1}{2}$.

In response to the update, write the factors $1 - 1/k^2$ as $(k-1)(k+1)/k^2$. Doing so, we observe that the product is equal to \begin{align}\prod_{j = 2}^n \frac{(j-1)(j+1)}{j^2} &= \frac{\prod_{j = 2}^n (j-1) \prod_{j = 2}^n (j + 1)}{\prod_{j = 2}^n j^2}\\ &= \frac{(n-1)!\frac{(n+1)!}{2}}{(n!)^2}\\
& = \frac{(n-1)!}{n!}\frac{(n+1)!}{2n!}\\
& = \frac{1}{n}\frac{n+1}{2} \\
&= \frac{n+1}{2n}\end{align}
The result is the same.

A:

Hint: use Euler's formula
$$\frac{\sin(\pi z)}{\pi z} = (1-z^2/1^2)(1-z^2/2^2)(1-z^2/3^2)\dots$$

