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?

$$\dfrac{n+1}{(n+5)(n+4)(n-3)} = -\dfrac12\cdot\dfrac1{n+5} + \dfrac37\cdot\dfrac1{n+4} + \dfrac1{14} \cdot\dfrac1{n-3}$$
Hence,
\begin{align}
\sum_{n=4}^{m} \dfrac{n+1}{(n+5)(n+4)(n-3)} & = - \dfrac12 \cdot \sum_{n=4}^{m} \dfrac1{n+5} + \dfrac37 \cdot \sum_{n=4}^{m} \dfrac1{n+4} + \dfrac1{14} \cdot \sum_{n=4}^{m} \dfrac1{n-3}\\
& = - \dfrac12 \sum_{n=9}^{m+5} \dfrac1n + \dfrac37 \sum_{n=8}^{m+4} \dfrac1n + \dfrac1{14} \sum_{n=1}^{m-3}\dfrac1n\\
& = - \dfrac12 \sum_{n=9}^{m+5} \dfrac1n + \dfrac37 \sum_{n=9}^{m+4} \dfrac1n + \dfrac1{14} \sum_{n=9}^{m-3}\dfrac1n + \dfrac37 \sum_{n=8}^{8} \dfrac1n + \dfrac1{14} \sum_{n=1}^{8}\dfrac1n\\
& = \dfrac{971}{3920} + \dfrac37 \sum_{m-2}^{m+4} \dfrac1n - \dfrac12 \sum_{m-2}^{m+5}\dfrac1n
\end{align}
Taking $m \to \infty$, we have $$\lim_{m \to \infty}\left(\dfrac37 \sum_{m-2}^{m+4} \dfrac1n - \dfrac12 \sum_{m-2}^{m+5}\dfrac1n \right) = 0$$
Hence,
$$\sum_{n=4}^{\infty} \dfrac{n+1}{(n+5)(n+4)(n-3)} = \dfrac{971}{3920}$$
