Sum of an Infinite Sequence 
If $$ S = \frac{1}{1\cdot3\cdot5} + \frac{1}{3\cdot 5 \cdot 7} +
  \frac{1}{5\cdot 7 \cdot 9} \cdots  $$
$$S =\, ? $$ 

My Attempt
Let the general term be $ a_n $.
Then, $$ a_n = \frac{1}{(2n-1)(2n+1)(2n+3)} $$
$$ a_n = \frac{1}{4} \left[\frac{(2n+3)-(2n-1)}{(2n-1)(2n+1)(2n+3)} \right]$$
$$\sum_{n=1}^{∞}a_n = \frac{1}{4} \left[\frac{1}{1\cdot3} - \frac{1}{(2n+1)(2n+3)}\right]$$ 
As n $\to$ ∞, 
 $$ \sum_{n=1}^{∞}a_n = \frac{1}{12}$$
Is this the correct?

You are right but more methodically:
For the series $$T_n=\frac{1}{(2n-1)(2n+1)(2n+3)}$$
Let $$V_n=\frac{1}{(2n-1)(2n+1)},$$ then $$T_n=\frac{1}{4}[V_{n}-V_{n+1}].$$ By telescopic summing we get $$S_N=\sum_{n=1}^{N}=\frac{1}{4} [V_1-V_{N+1}]=\frac{1}{4} \left[\frac{1}{1.3}-\frac{1}{(2N+1)(2N+3)}\right].$$ Hence $$S_{\infty}=\frac{1}{12}.$$
