Conversion of sum of series into product form: $\sum_{n=1}^\infty \frac1{n(n+1)} = \frac12 \prod_{n=2}^\infty \left( 1+\frac{1}{n^2-1} \right)$ Show that the following series and product are equivalent:
$$
\sum_{n=1}^\infty \left[ \dfrac{1}{n(n+1)} \right] = \dfrac{1}{2}  \prod_{n=2}^\infty \left[ 1+\dfrac{1}{n^2-1} \right]
$$
Thought of solving this through induction as the relation is true for 1....but don't know how to proceed with the n part...Please help!!

Hint: First, simplify the fractions to get
$$\prod\limits_{n = 2}^\infty 1 + \frac{1}{n^2 - 1} = \prod\limits_{n = 2}^\infty \frac{n^2}{n^2 - 1}$$
Denote the partial product by $P_n$, and take a logarithm:
$$\log{P_N} = \log{\prod\limits_{n = 2}^N \frac{n^2}{n^2 - 1}} = \sum\limits_{n = 2}^N \log{\frac{n^2}{n^2 - 1}}$$
$$ = \sum\limits_{n = 2}^N \log{n^2} - \log{(n^2 - 1)} = \sum\limits_{n = 2}^N 2\log(n) - \log(n - 1) - \log(n + 1)$$
Now a lot of terms will telescope.
