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.
 A: Note that $$n^3+6n^2+8n = n(n+4)(n+2)$$
Thus you need to apply partial fraction on $$ \frac {1}{n(n+4)(n+2)} $$and make it some sort of telescoping series to find the exact value.
A: Note
$$ \frac{1}{n^3+6n^2+8n}=\frac1{n(n+2)(n+4)}=\frac14\bigg[\frac{1}{n(n+2)}-\frac{1}{(n+2)(n+4)}\bigg]. $$
Let
$$ f(x)=\sum_{n=1}^\infty\frac{1}{n(n+2)}x^n, g(x)=\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)}x^{n+2}. $$
Then
\begin{eqnarray}
f(1)&=&\int_0^1\frac1{x^3}\int_0^x\frac{t^2}{1-t}dtdx\\
&=&\int_0^1\int_0^x\frac{t^4}{x^2(1-t)}dtdx\\
&=&\int_0^1\int_t^1\frac{t^4}{x^2(1-t)}dxdt\\
&=&\frac12\int_0^1\frac{t^2}{1-t}\bigg(\frac1{t^2}-1\bigg)dt\\
&=&\frac12\int_0^1(1+t)dt\\
&=&\frac{3}{4},\\
g(1)&=&\int_0^1\frac1{x^3}\int_0^x\frac{t^4}{1-t}dtdx\\
&=&\int_0^1\int_0^x\frac{t^4}{x^3(1-t)}dtdx\\
&=&\int_0^1\int_t^1\frac{t^4}{x^3(1-t)}dxdt\\
&=&\frac12\int_0^1\frac{t^4}{1-t}\bigg(\frac1{t^2}-1\bigg)dt\\
&=&\frac12\int_0^1t^2(1+t)dt\\
&=&\frac{7}{24}.
\end{eqnarray}
So
$$ \sum_{n=1}^\infty\frac1{n(n+2)(n+4)}=\frac14\bigg(\frac23-\frac{7}{24}\bigg)=\frac{11}{96}.$$
A: 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.
