Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the 
$$\sum\limits_{n = 4}^\infty  {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$
series I used the criterion of integral $$\displaystyle\int_4^\infty  {\frac{{x + 1}}{{(x + 5)(x + 4)(x - 3)}}dx},$$ but calculate this improper integral is a very laborious task. 
Is there a shorter way? What criteria of convergence would be most effective or simple?

In the same spirit as Leg's answer, starting with the partial fraction decomposition $$\dfrac{n+1}{(n+5)(n+4)(n-3)} = -\dfrac12\cdot\dfrac1{n+5} + \dfrac37\cdot\dfrac1{n+4} + \dfrac1{14} \cdot\dfrac1{n-3}$$ and using harmonic numbers definition $$\sum_{n=p}^m \dfrac1{n+5}=H_{m+5}-H_{p+4}$$ $$\sum_{n=p}^m \dfrac1{n+4}=H_{m+4}-H_{p+3}$$ $$\sum_{n=p}^m \dfrac1{n-3}=H_{m-3}-H_{p-4}$$ which make $$A_p=\sum_{n=p}^m \dfrac{n+1}{(n+5)(n+4)(n-3)} =\frac{1}{14} \left(H_{m-3}+6 H_{m+4}-7 H_{m+5}-H_{p-4}+H_{p+3}+\frac{7}{p+4}\right)$$ When $m\to \infty$, $H_{m-3}+6 H_{m+4}-7 H_{m+5}\to 0$ and then, at the limit, $$A_p=\frac{1}{14}
   \left(\frac{1}{p-3}+\frac{1}{p-2}+\frac{1}{p-1}+\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}
   {p+3}+\frac{7}{p+4}\right)$$ which gives $$A_4=\frac{971}{3920}$$ $$A_5=\frac{6289}{35280}$$ $$A_6=\frac{5113}{35280}$$ as so forth.
