<s>[INST] 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
$$(2^23^24^2...n^2) = (n!)^2$$?
 [/INST] 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 $x=1$.
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