Finding the sum of a series that converges

Question

I have a series $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3+6n^2+8n}$$

I know the series converges because $\frac{1}{n^3+6n^2+8n}$ is less than or equal to $1/n^3$. Since $p=3$ which is greater than $1$, I know that $1/n^3$ converges. Since that converges I also know $\frac{1}{n^3+6n^2+8n}$ converges, but I am not sure how to figure out what it converges to.

Answer 1

Note that $n^3+6n^2+8n=n(n+2)(n+4)$ hence, by partial fraction decomposition, $$\frac{1}{n^3+6n^2+8n}=\frac{A}{n(n+2)}+\frac{B}{(n+2)(n+4)}$$ for some constants $A$ and $B$ to be found.

Moreover, by letting $m=n+2$, $$\sum_{n=1}^{\infty}\frac{1}{(n+2)(n+4)}=\sum_{m=3}^{\infty}\frac{1}{m(m+2)}.$$

Can you take it from here?

Answer 2

I used a different partial fraction decomposition: $n^3+6n^2+8n = n(n+2)(n+4)$

$\dfrac{1}{n^3+6n^2+8n} = \dfrac{1}{8n}+\dfrac{1}{8(n+4)}-\dfrac{2}{8(n+2)}$

From here, I set up a table:

$\begin{matrix} & \dfrac{1}{8} & \dfrac{1}{8(2)} & \dfrac{1}{8(3)} & \dfrac{1}{8(4)} & \dfrac{1}{8(5)} & \dfrac{1}{8(6)} & \dfrac{1}{8(7)} & \dfrac{1}{8(8)} & \cdots \\ \dfrac{1}{8n} & 1 & 1 & 1 & 1& 1 & 1 & 1 & 1 & \cdots \\ \dfrac{1}{8(n+4)} & & & & & 1 & 1 & 1 & 1 & \cdots \\ \dfrac{-2}{8(n+2)} & & & -2 & -2 & -2 & -2 & -2 & -2 & \cdots \\ \text{Total} & 1 & 1 & -1 & -1 & & & & & \cdots\end{matrix}$