Calculate $\sum\limits_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$ 
Calculate the sum of the series $$\sum_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$$

I have tried partial fraction decomposition.
$$\sum_{n=1}^\infty\frac{1}{4(n+2)}- \sum_{n=1}^\infty\frac{1}{4(n+4)}-\sum_{n=1}^\infty\frac{1}{2(n+4)^2}$$
Is this correct?  What is the sum?

First note that your series has the same convergence as $\sum \frac{1}{n^3}$ by the limit comparison test. And the latter series converges absolutely by the $p$-series test or the integral test, so therefore so does your series.
Next, to find the value the sum converges to, trying partial fractions, we get
$$\frac{1}{(n+2)(n+4)(n+4)}=\frac{1}{4}\frac{1}{n+2}-\frac{1}{4}\frac{1}{n+4}-\frac{1}{2}\frac{1}{(n+4)^2}$$
If we break it into three series as you have attempted, then all three are divergent, and difference of divergent is indeterminate, so we cannot proceed. Instead, treat the first two terms together, and note that $\sum_{n=1}^\infty \frac{1}{n+2}-\frac{1}{n+4}$ is a telescoping series, its sum will converge to $1/3+1/4$, the uncanceled parts of the first two terms. For the final term, rememeber by the Basel problem $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ so by reindexing we have $\sum_{n=5}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{(n+4)^2}=\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1$.
Putting it all together we have
$$\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)(n+4)}=\frac{1}{4}\left(\sum_{n=1}^\infty\frac{1}{n+2}-\frac{1}{n+4}\right)-\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(n+4)^2} \\=\frac{1}{4}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{2}\left(\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1\right) = \frac{7}{48} -\frac{\pi^2}{12}+\frac{205}{288}=\frac{247}{288}-\frac{\pi^2}{12}
$$ 
