(a) $-\frac{1}{x+2}+\frac{3}{x+3}-\frac{2}{x+4}$

(b) $\frac{1}{6}+\frac{1}{n+3}-\frac{2}{n+4}$

(c) $\frac{1}{6}$

(a) Let $\frac{x}{(x+2)(x+3)(x+4)}=\frac{A}{x+2}+\frac{B}{x+3}+\frac{C}{x+4}$ $x=A(x+3)(x+4)+B(x+2)(x+4)+$ $C(x+2)(x+3)$
Let $x=-3$,
$-3=B(-1)(1) \quad \Rightarrow B=3$
Let $x=-2$,
$-2=A(1)(2) \quad \Rightarrow A=-1$
Let $x=-4$,
$$
-4=C(-2)(-1) \Rightarrow C=-2
$$
$$
\therefore \frac{x}{(x+2)(x+3)(x+4)}=-\frac{1}{x+2}+\frac{3}{x+3}-\frac{2}{x+4}
$$

$\sum_{r=1}^n \frac{r}{(r+2)(r+3)(r+4)}$
$=\sum_{r-1}^n\left(-\frac{1}{r+2}+\frac{3}{r+3}-\frac{2}{r+4}\right)$

$=-\frac{1}{3}+\frac{3}{4}-\frac{1}{4}-\frac{2}{n+3}+\frac{3}{n+3}-\frac{2}{n+4}$
$=\frac{1}{6}+\frac{1}{n+3}-\frac{2}{n+4}$

(c) As $n \rightarrow \infty, \frac{1}{n+3} \rightarrow 0$ and $\frac{2}{n+4} \rightarrow 0$
$\therefore \sum_{r=1}^{\infty} \frac{r}{(r+2)(r+3)(r+4)}=\frac{1}{6}$