Exact value of a convergent series: $\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}$ I have a series
$$\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}.$$
I know the series converges because 
$$\frac{1}{n^3+6n^2+8n}\le \frac{1}{n^3}, $$
since $p=3>1$, I know that $\sum 1/n^3$ converges. But I am not sure how to figure out what it converges to.

Just another way using partial sums.
Using Mohammad Riazi-Kermani's answer, use partial fraction decomposition to get
$$a_n=\frac 1{n^3+6n^2+8n}=\frac {1}{n(n+4)(n+2)}=-\frac{1}{4 (n+2)}+\frac{1}{8 (n+4)}+\frac{1}{8 n}$$ and consider
$$S_p=\sum_{n=1}^p a_n$$ Using harmonic numbers
$$\sum_{n=1}^p \frac 1n=H_p\qquad \sum_{n=1}^p \frac 1{n+2}=H_{p+2}-\frac{3}{2}\qquad \sum_{n=1}^p \frac 1{n+4}=H_{p+4}-\frac{25}{12}$$
Now, use the asymptotics
$$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12
   q^2}+O\left(\frac{1}{q^3}\right)$$ Replace and continue with Taylor expansions for large $p$ to get 
$$S_p=\frac{11}{96}-\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and also how it is approached.
