How do I evaluate the limit,
$$\lim_{n \to \infty} \ \ \ \left(1-\frac1{2^2}\right)\left(1-\frac1{3^2}\right)\left(1-\frac1{4^2}\right)...\left(1-\frac1{n^2}\right)$$
I tried to break the $n^{\text{th}}$ term into $$\frac{(n+1)(n-1)}{n.n}$$ and then tried to do something but couldn't get anywhere. Please help as to how to solve such questions in general. Thanks.
On
Did you try starting with small sequences and then seeing what cancels out each time?
On
Hint.
$$ \frac{(2-1)(2+1)}{2\cdot 2}\cdots \frac{(n-2)n}{(n-1)(n-1)}\frac{(n-1)(n+1)}{nn}\frac{n(n+2)}{(n+1)(n+1)}\frac{(n+1)(n+3)}{(n+2)(n+2)} = \frac 12\frac{(n+3)}{(n+2)} $$
On
The following will not be of help for such questions "in general", but it is too long for a comment.
Following Euler, we have $$ \begin{align} \frac{\sin x}{x} &= \left( 1 - \frac{x}{\pi} \right) \left( 1 + \frac{x}{\pi} \right) \left( 1 - \frac{x}{2 \pi} \right) \left( 1 + \frac{x}{2 \pi} \right) \left( 1 - \frac{x}{3 \pi} \right) \left( 1 + \frac{x}{3 \pi} \right) \dots \\ &= \left[ 1 - \left( \frac{x}{\pi} \right)^2 \right] \left[ 1 - \frac{1}{2^2} \left( \frac{x}{\pi} \right)^2 \right] \left[ 1 - \frac{1}{3^2} \left( \frac{x}{\pi} \right)^2 \right] \dots, \end{align} $$ or $$ \frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2} = \left[ 1 - \frac{1}{2^2} \left( \frac{x}{\pi} \right)^2 \right] \left[ 1 - \frac{1}{3^2} \left( \frac{x}{\pi} \right)^2 \right] \dots. $$ Hence the required product is $$ \begin{align} \lim_{x \to \pi} \frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2} &= \lim_{\epsilon \to 0} \frac{\sin \epsilon}{\pi - \epsilon} \cdot \frac{1}{1 - ([\pi - \epsilon]/\pi)^2} \\ &= \lim_{\epsilon \to 0} \frac{\epsilon}{\pi} \cdot \frac{1}{2 \pi \epsilon / \pi^2} \\ &= \frac{1}{2}. \end{align} $$
We have $$ \log\left(\prod_{n =2}^N 1 - \frac{1}{n^2} \right) = \sum_{n = 2}^N \log(n-1) + \log(n+1) - 2 \log(n) = \log(1) - \log(2) - \log(N) + \log(N+1).$$ So we have that $$\prod_{n = 1}^N 1 - \frac{1}{n^2} = \frac{N+1}{2N} \to \frac{1}{2}$$