Series sum Diverge or Converge??

Question

\begin{equation} \sum_{n=1}^{\infty}\frac{n-1}{(n+2)(n+3)} \end{equation}

The answer was Divergent, but I solved it as Convergent by using limit comparison test. plz help me solving this question.

Answer 1

Use partial fraction decomposition to rewrite $$\frac{n-1}{(n+2)(n+3)}=\frac{4}{n+3}-\frac{3}{n+2}=\frac{1}{n+3}+3\left(\frac{1}{n+3}-\frac{1}{n+2 } \right)$$

$$\sum_{n=1}^\infty\frac{n-1}{(n+2)(n+3)}=\sum_{n=1}^\infty\frac{1}{n+3}+3\sum_{n=1}^\infty \left(\frac{1}{n+3}-\frac{1}{n+2 } \right)$$ The second summation telescopes and then $$\sum_{n=1}^\infty\frac{n-1}{(n+2)(n+3)}=\sum_{n=1}^\infty\frac{1}{n+3}-1$$

Answer 2

$(n+2)(n+3) = n^{2} + 5n + 6 \leq n^{2} + 5n^{2} + 6n^{2} \leq 13n^{2}$

And $n-1 \geq \frac{n}{2}$ for all $n > 1$

Then $\frac{n-1}{(n+2)(n+3)} \geq \frac{\frac{n}{2}}{13n^{2}} = \frac{1}{26n}$

Thus $\sum_{n = 2}^{\infty} \frac{n-1}{(n+2)(n+3)} \geq \sum_{n=2}^{\infty} \frac{1}{26n} = \frac{1}{26} \sum_{n=2}^{\infty} \frac{1}{n} = \infty$

Then by comparison $\sum_{n = 2}^{\infty} \frac{n-1}{(n+2)(n+3)}$ is divergent $\implies \sum_{n = 1}^{\infty} \frac{n-1}{(n+2)(n+3)}$ is divergent

Answer 3

$$ \left. \frac{n-1}{(n+2)(n+3)} \, \right/ \frac 1 n \to 1 \text{ as } n\to\infty. $$ Therefore $$ \left. \frac{n-1}{(n+2)(n+3)} \, \right/ \frac 1 n > 0.9 \text{ for large enough values of } n. $$ So $$ \frac{n-1}{(n+2)(n+3)} > \frac{0.9} n \text{ for large enough values of }n. $$ So by a simple comparison test, the series considered diverges to $+\infty.$