Calculate the sum of the series $\sum_{n=1}^{\infty} \frac{n+12}{n^3+5n^2+6n}$ They tell me to find the sum of the series
$$\sum a_n :=\sum_{n=1}^{\infty}\frac{n+12}{n^3+5n^2+6n}$$
Since $\sum a_n$ is absolutely convergent, hence we can manipulate it the same way we would do with finite sums. I've tried splitting the general term and I get
$$\frac{n+12}{n(n+2)(n+3)}=\frac{A}{n}+\frac{B}{n+2}+\frac{C}{n+3}=\frac{2}{n}+\frac{-5}{n+2}+\frac{3}{n+3}$$
and so
$$\sum a_n =\sum \frac{2}{n}-\frac{5}{n+2}+\frac{3}{n+3}$$
Now, if I was to split the series in the sum of three different series I would get three different divergent series and so, obviously $\sum a_n$ wouldn't converge.
I also suspect about being a telescopic series althought the numerators of each fraction makes it difficult to find the cancellation terms. I also know that I can rearrange the terms in my series, althought I cannot see how would this solve the problem.
If anyone could give me a hint I would really appreciate it.

If you know how to play with harmonic numbers, it is pretty simple
$$\sum_{n=1}^p \frac 1 {n}=H_p$$
$$\sum_{n=1}^p \frac 1 {n+2}=H_{p+2}-\frac{3}{2}$$
$$\sum_{n=1}^p \frac 1 {n+3}=H_{p+3}-\frac{11}{6}$$
Now, use three times the asymptotics
$$H_q=\log (q)+\gamma +\frac{1}{2 q}-\frac{1}{12 q^2}+O\left(\frac{1}{q^4}\right)$$ and continue with Taylor or long division and you will have
$$\sum_{n=1}^{p}\frac{n+12}{n^3+5n^2+6n}=2-\frac{1}{p}-\frac{3}{p^2}+O\left(\frac{1}{p^3}\right)$$
Use it for $p=10$; the approximation gives exactly $1.87$ while the exact value is $\frac{1615}{858}=1.88228$ so a relative error of $0.65$%.
