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?
 A: Hint, ignore the constants and this is the series:
$$\sum\limits_{n=1}^{\infty} \frac{n}{n^3}=\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$$
which is convergent. 
More formally, use the limit comparison test with $\frac{1}{n^2}$.
A: Assume that the series begins at $n=4$.  Then, we have 
$$n+5\ge n$$
$$n+4\ge n$$
$$n-3\ge \frac14 n$$
$$n+1\le 2n$$
Therefore, have 
$$\frac{n+1}{(n+5)(n+4)(n-3)}\le \frac{2n}{\frac14 n^3}=8\frac1{n^2}$$
Finally, using the result $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ reveals
$$\sum_{n=4}^{\infty}\frac{n+1}{(n+5)(n+4)(n-3)}\le 8\sum_{n=4}^{\infty}\frac1{n^2}=4\left(\frac{\pi^2}{3}-3\right)$$
and the series converges by the comparison test.
A: $$\dfrac{n+1}{(n+5)(n+4)(n-3)} = -\dfrac12\cdot\dfrac1{n+5} + \dfrac37\cdot\dfrac1{n+4} + \dfrac1{14} \cdot\dfrac1{n-3}$$
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,
$$\sum_{n=4}^{\infty} \dfrac{n+1}{(n+5)(n+4)(n-3)} = \dfrac{971}{3920}$$
A: 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.
