What is the limit of $\lim_{n\to \infty} (1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16}) \cdots (1 - \frac{1}{(n+1)^2})$? What is the evaluation of the following infinite series?
$$\lim_{n\to \infty} \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right) \cdots \left(1- \frac{1}{(n+1)^2}\right)$$
I've tried to simplify each expression which left me with:
$$\lim_{n\to \infty} \frac{3\times8\times15\times24\times\cdots\times((n+1)^2-1)}{4\times9\times16\times25\times\cdots\times(n+1)^2}$$
Is this a good way to approach this problem? 

You want
\begin{align*}\prod_{k=1}^{n} \left(1-\frac{1}{(k+1)^2}\right)
& = \prod_{k=1}^{n} \left(1-\frac{1}{k+1}\right)\left(1+\frac{1}{k+1}\right) \\
& = \prod_{k=1}^{n} \left(1-\frac{1}{k+1}\right)\prod_{k=1}^{n} \left(1+\frac{1}{k+1}\right) \tag{1}\end{align*}
Now,
$$
\prod_{k=1}^{n} \left(1-\frac{1}{k+1}\right) = \prod_{k=1}^{n} \frac{k}{k+1} = \frac{1}{n+1}\tag{2}
$$
(as things cancel out/telescope), and similarly
$$
\prod_{k=1}^{n} \left(1+\frac{1}{k+1}\right) = \prod_{k=1}^{n} \frac{k+2}{k+1} = \frac{n+2}{2} \tag{3}
$$ 
and so, combining (2) and (3) into (1),
$$
\prod_{k=1}^{n} \left(1-\frac{1}{(k+1)^2}\right) = \frac{1}{2}\cdot\frac{n+2}{n+1}
$$
