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? 
 A: 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),
$$
\prod_{k=1}^{n} \left(1-\frac{1}{(k+1)^2}\right) = \frac{1}{2}\cdot\frac{n+2}{n+1}
$$
A: $$\prod_{k=1}^n\dfrac{f(k)}{f(k+1)}=\dfrac{f(1)}{f(n+1)}$$
Here $f(m)=\dfrac m{m+1}$
A: $\frac{1\times 3}{2\times 2}\cdot\frac{2\times 4}{3\times 3}\cdot\frac{3\times 5}{4\times 4}\cdot\frac{4\times 6}{5\times 5}\cdots$
Do you see it?
A: hint
Take logarithm and use
$$
\sum_{k=2}^{n+1}\ln\left(1-\frac{1}{k^2}\right)
= \sum_{k=2}^{n+1}\left(\ln\left(\frac{k-1}{k}\right)+\ln\left(\frac{k+1}{k}\right)\right)
=\ln\left(\frac{1}{n+1}\right)+\ln\left(\frac{n+2}{2}\right).
$$
A: Since 
$$\frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right) $$
we simply have
$$ \prod_{n\geq 2}\left(1-\frac{1}{n^2}\right)=\lim_{x\to 1}\frac{\sin(\pi x)}{\pi x(1-x^2)}=\lim_{z\to 0}\frac{-\sin(\pi z)}{-\pi z(z+1)(z+2)}=\color{red}{\frac{1}{2}}. $$
This technique is interesting since it allows a quick evaluation of $\prod_{n\geq 1}\left(1-\frac{1}{2n^2}\right)$ and similar products, which are not telescopic.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\prod_{k = 1}^{n}\bracks{1 - {1 \over \pars{k + 1}^{2}}}} =
\overbrace{{\prod_{k = 1}^{n}k \over \prod_{k = 1}^{n}\pars{k + 1}}}^{\ds{=\ {1 \over n + 1}}}\
\overbrace{{\prod_{k = 1}^{n}\pars{k + 2} \over
\prod_{k = 1}^{n}\pars{k + 1}}}^{\ds{=\ {n + 2 \over 2}}} \\[5mm] = &\
{1 \over 2}\,{1 + 2/n \over 1 + 1/n}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}
{\Large \to}\,\,\,\bbx{1 \over 2}
\end{align}
