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 $\dfrac{11}{96}$
 A: HINT you need to do a partial fraction decomposition. When you start writing out the terms you realise you have terms cancelling (it's a telescoping series). I would write it out for you but it's more fun for you to see it for yourself. :)
A: 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:
$$\frac18\biggl(1+\frac12-\frac13-\frac14-\frac1{n+1}-\frac1{n+2}+\frac1{n+3}+\frac1{n+4}\biggr).$$
A: This might be overkill but,\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n(n+2)(n+4)}
&=\frac18\sum_{n=1}^{\infty}\left(\frac{1}{n+4}+\frac{1}{ n}
-\frac{2}{n+2}\right)\tag{1}\\
&=\frac18\sum_{n=1}^{\infty}\int_{0}^{1}\left(x^{n+3}+x^{n-1}-2x^{n+1}\right)\,\mathrm dx\tag{2}\\
&=\frac18\int_{0}^{1}\sum_{n=1}^{\infty}\left(x^{n+3}+x^{n-1}-2x^{n+1}\right)\,\mathrm dx\tag{3}\\
&=\frac{1}{8}\int_{0}^{1}\left(\frac{x^{4}}{1-x}+\frac{1}{1-x}-\frac{2x^2}{1-x}\right)\,\mathrm dx\tag{4}\\
&=\frac18\int_{0}^{1}\frac{(x^2-1)^2}{1-x}\,\mathrm dx\tag{5}\\
&=\frac18\int_{0}^{1}\left(1+x-x^2-x^3\right) \,\mathrm dx\tag{6}\\
&=\frac18\Bigg[x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}\Bigg]_{0}^{1}\tag{7}\\
&=\frac{11}{96}\tag{8}\\
\end{align}

$$\sum_{n=1}^{\infty}\frac{1}{n(n+2)(n+4)}=\frac{11}{96}$$

A: HINT:
$$\displaystyle\dfrac4{n(n+2)(n+4)}=\dfrac{n+4-n}{n(n+2)(n+4)}=\dfrac1{n(n+2)}-\dfrac1{(n+2)(n+4)}$$
$$=g(n)-g(n+2)$$
where $g(m)=\dfrac1{m(m+2)}$
See Telescoping series
A: $$ \frac{1}{n(n+2)(n+4)}=\frac{1}{8}(\frac{1}{n}-\frac{1}{n+2})-\frac{1}{8}(\frac{1}{(n+2}-\frac{1}{n+4})$$
we know 
$$\sum_{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\sum_{n=1}^{\infty }\frac{1}{n+2}$$
$$\sum_{n=1}^{\infty }\frac{1}{n+2}=\frac{1}{3}+\frac{1}{4}+\sum_{n=1}^{\infty }\frac{1}{n+4}$$
$$\frac{1}{8}(\frac{1}{1}+\frac{1}{2})-\frac{1}{8}(\frac{1}{3}+\frac{1}{4})=\frac{11}{96}$$
