Find the limit of a multiplying term function when n tends to infinity. 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.
 A: Hint: Prove by induction that $$\prod_{i=2}^n 1-\frac{1}{i^2}=\frac{n+1}{2n}$$
A: Hint.
$$
\frac{(2-1)(2+1)}{2\cdot 2}\cdots \frac{(n-2)n}{(n-1)(n-1)}\frac{(n-1)(n+1)}{nn}\frac{n(n+2)}{(n+1)(n+1)}\frac{(n+1)(n+3)}{(n+2)(n+2)} = \frac 12\frac{(n+3)}{(n+2)}
$$
A: 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
$$\prod_{n = 1}^N 1 - \frac{1}{n^2} = \frac{N+1}{2N} \to \frac{1}{2}$$
A: Did you try starting with small sequences and then seeing what cancels out each time?


*

*$\frac{1*3}{2*2} = \frac{3}{4}$

*$\frac{3}{4} * \frac{2*4}{3*3} = \frac{2}{3}$

*$\frac{2}{3} * \frac{3*5}{4*4} = \frac{5}{8}$

*$\frac{5}{8} * \frac{4*6}{5*5} = \frac{3}{5}$
A: 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
$$
\begin{align}
\lim_{x \to \pi} \frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2}
&= \lim_{\epsilon \to 0} \frac{\sin \epsilon}{\pi - \epsilon} \cdot \frac{1}{1 - ([\pi - \epsilon]/\pi)^2} \\
&= \lim_{\epsilon \to 0} \frac{\epsilon}{\pi} \cdot \frac{1}{2 \pi \epsilon / \pi^2} \\
&= \frac{1}{2}.
\end{align}
$$
