sequences and series - Evaluate $sum_{n=1}^infty frac{1}{n(n+2)(n+4)}$?

How can I evaluate this?

$$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)} = \frac{1}{1\cdot3\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{3\cdot5\cdot7}+ \frac{1}{4\cdot6\cdot8}+\cdots$$

I have tried:

$$\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{2\cdot4\cdot6}+ \frac{1}{4\cdot6\cdot8}+\cdots = \frac{1}{3\cdot5}\left(1+\frac{1}{7}\right)+\frac{1}{4\cdot6}\left(\frac{1}{2}+\frac{1}{8}\right)+\cdots$$

and so on... Been stuck for a while. Result should be $\dfrac{11}{96}$

Answer

With *partial fractions: $$\frac1{k(k+2)(k+4)}=\frac18\biggl(\frac1k-\frac2{k+2}+\frac1{k+4}\biggr),$$ we get a telescoping sum which simplifies to: $$\frac18\biggl(1+\frac12-\frac13-\frac14-\frac1{n+1}-\frac1{n+2}+\frac1{n+3}+\frac1{n+4}\biggr).$$