Sum of the series $\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$. Sum to n terms and also to infinity of the following series:

$$\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$$

the solution provided by the book is 
$$S_n=\frac{17}{96}-[\frac{1}{n+5}+\frac{1}{2(n+4)(n+5)}-\frac{1}{2(n+3)(n+4)}-\frac{3}{4(n+2)(n+3)(n+4)(n+5)}]$$
And $S_\infty=\dfrac{17}{96}$.Can anyone help me to explain how to get $S_n$.
Thanks in Advanced.
 A: $$S_n=\dfrac{n}{(n+1)(n+3)(n+5)}=\dfrac{A}{n+1}+\dfrac{B}{n+3}+\dfrac{C}{n+5}$$
then $A=-\dfrac18$, $B=\dfrac68$ and $C=-\dfrac58$ then
$$S_n=\dfrac{n}{(n+1)(n+3)(n+5)}=\dfrac18\left(\dfrac{1}{n+3}-\dfrac{1}{n+1}\right) - \dfrac58\left(\dfrac{1}{n+5}-\dfrac{1}{n+3}\right)$$
for using telescopic method we need add some terms:
$$S_n=\dfrac18\left(\dfrac{1}{n+3}\color{red}{-\dfrac{1}{n+2}}\right) + \dfrac18\left(\color{red}{\dfrac{1}{n+2}}-\dfrac{1}{n+1}\right) -\dfrac58 \left(\dfrac{1}{n+5}\color{red}{-\dfrac{1}{n+4}}\right) - \dfrac58\left(\color{red}{\dfrac{1}{n+4}}-\dfrac{1}{n+3}\right)$$
telescopic method says $\displaystyle\sum_{k=p}^q a_{k+1}-a_k=a_{q+1}-a_p$ therefore
$$S_n=\dfrac18\dfrac{1}{n+3} + \dfrac18\dfrac{1}{n+2} - \dfrac58 \dfrac{1}{n+5} - \dfrac58\dfrac{1}{n+4} + \dfrac{17}{96}$$
and $\color{blue}{S_\infty=\dfrac{17}{96}}$.
A: Alternatively, write
$$
\begin{align}\frac{n}{(n+1)(n+3)(n+5)}&=\frac{(n+5)-5}{(n+1)(n+3)(n+5)}\\&=\frac{1}{(n+1)(n+3)}-\frac{5}{(n+1)(n+3)(n+5)}\,.\end{align}$$
Now,
$$\frac{1}{(n+1)(n+3)}=\frac{1}{2}\,\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$$
and
$$\frac{1}{(n+1)(n+3)(n+5)}=\frac{1}{4}\,\left(\frac{1}{(n+1)(n+3)}-\frac{1}{(n+3)(n+5)}\right)\,.$$
That is,
$$\frac{n}{(n+1)(n+3)(n+5)}=\frac{1}{2}\,\left(\frac{1}{n+1}-\frac{1}{n+3}\right)-\frac{5}{4}\,\left(\frac{1}{(n+1)(n+3)}-\frac{1}{(n+3)(n+5)}\right)\,.$$
The rest is straightforward.
