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.

Solution 1:

As you said, for every $N \in \mathbb{N}^*$, one has \begin{align*}\sum_{n=1}^N a_n &=\sum_{n=1}^N \frac{2}{n}-\frac{5}{n+2}+\frac{3}{n+3}\\ &=2 \sum_{n=1}^N \frac{1}{n} - 5 \sum_{n=3}^{N+2} \frac{1}{n} + 3 \sum_{n=4}^{N+3} \frac{1}{n}\\ &=2 \left(1 + \frac{1}{2} + \frac{1}{3} + \sum_{n=4}^N \frac{1}{n} \right) - 5 \left( \frac{1}{3} + \sum_{n=4}^N \frac{1}{n} + \frac{1}{N+1} + \frac{1}{N+2} \right)\\ &\quad + 3 \left( \sum_{n=4}^N \frac{1}{n} + \frac{1}{N+1} + \frac{1}{N+2} + \frac{1}{N+3} \right)\\ &= 2 - 5 \left( \frac{1}{N+1} + \frac{1}{N+2} \right) + 3 \left( \frac{1}{N+1} + \frac{1}{N+2} + \frac{1}{N+3} \right) \end{align*}

Now just let $N$ tend to $+\infty$ to see that $$\boxed{\sum_{n=1}^{\infty} a_n = 2}$$