<s>[INST] Lim of $a_{n} = (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$ $a_{n} = (1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})$
I first find that $a_{n}$ is descending
$$\frac{a_{n+1}}{a_{n}}=\cfrac{(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})(1-\frac{1}{(n+1)^2})}{(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{n^2})} = 1-\frac{1}{(n+1)^2}<1$$
So $a_{n}$ is descending and also it is $$0< 1-\frac{1}{(n+1)^2}<1$$
So I can say $$\exists\lim\limits_{n \to \infty} a_{n} =l\in\Bbb{R} $$
How do i continue?
 [/INST] We have
$$a_n = \dfrac{1 \cdot 3}{2\cdot2} \cdot \dfrac{2 \cdot 4}{3\cdot3} \cdot \dfrac{3 \cdot 5}{4\cdot4}  \cdot \dfrac{4 \cdot 6}{5\cdot5} \cdots \dfrac{(n-1)\cdot (n+1)}{n \cdot n}$$
Telescopic cancellation shows that
$$a_n = \dfrac12 \cdot \dfrac{(n+1)}{n},\,\, \forall n \geq 2$$
You can also check this by induction. Now it is easy to see what $\lim_{n \to \infty} a_n$ is.
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