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!!
 A: $\frac 1{n(n+1)}= \frac1n-\frac1{n+1}$ so this is a telescoping sum, almost all terms will cancel out:
$$ \sum_{n=1}^N \frac1{n(n+1)}= 1 - \frac12 + \frac12 - \frac13 + \dots + \frac1{N-1}-\frac1N + \frac1N-\frac1{N+1}=1-\frac1{N+1}.$$
Similarly, you can see that in the product of the terms of the form $\frac{n^2}{n^2-1}=\frac{n^2}{(n+1)(n-1)}$ a lot of terms cancel out, you only need to check which of them.
$$\prod_{n=2}^N \frac{n^2}{(n+1)(n-1)} = \frac{2^2}{3\cdot 1} \cdot \frac{3^2}{2\cdot 4} \dots \frac{(N-1)^2}{(N-2)N} \cdot \frac{N^2}{(N+1)(N-1)}=
2 \cdot \frac{N}{N+1}.$$ 
Now it only remains to look what happens for $N\to\infty$ or to notice that already partial sums are the same as partial products, as pointed out in achille hui's comment.
A: 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.
