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: Hint: use Euler's formula
$$\frac{\sin(\pi z)}{\pi z} = (1-z^2/1^2)(1-z^2/2^2)(1-z^2/3^2)\dots$$
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: The function $sin(x)$ has Zeros at $..., - \pi, 0, \pi, 2 \pi, ...$. Therefore one can write:
$sin(\pi x) = ...(x+2 )(x+1)(x)(x-1)(x-2 )... = ...(x^2-2^2 )x(x^2-1^2)...$.
Rearrange this Expression by dividing through $1^22^2...$ and then Substitute $x=1$.
A: Write the product as $P= e^{ \log P}$. Using the property of logarithm you get 
$$
e^{\sum_{k=1}^{n}\log (k+1) + \sum_{k=2}^{n} \log (k-1) - 2 \sum_{k=1}^{n} \log k}
$$
Now either use Euler-Maclaurin formula or just the bounds on the sum: $\int_{1}^{n} \log x dx < \sum_{k=1}^{n} \log k < \int_{1}^{n+1} \log x dx$ and keep in miond that terms like $\frac{1}{n} \to e^{\frac{1}{n}} \to 1$
