(i) $\frac{1}{2}-\frac{1}{(n+2) !}$  $A=-1, B=1$   

(ii) 1

(i) $$\frac{1}{r !(r+2)}=\frac{r+1}{(r+2) !}=\frac{r+2-1}{(r+2) !}=\frac{1}{(r+1) !}-\frac{1}{(r+2) !}$$
Alternative:
$$\frac{1}{r !(r+2)}=\frac{r+1}{(r+2) !}$$
$$\begin{array}{l}
\frac{A}{(r+2) !}+\frac{B}{(r+1) !}=\frac{A+B(r+2)}{(r+2) !} \\
\therefore r+1=A+B(r+2)
\end{array}$$
By comparing coeff: $A=-1, B=1$

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

(ii) $$
\sum_{r=1}^{\infty} \frac{1+\left(\frac{1}{3}\right)^r r !(r+2)}{r !(r+2)}=\sum_{r=1}^{\infty} \frac{1}{r !(r+2)}+\sum_{r=1}^{\infty}\left(\frac{1}{3}\right)^r
$$As $n \rightarrow \infty, \frac{1}{(n+2) !} \rightarrow \infty$, thus $\sum_{r=1}^{\infty} \frac{1}{r !(r+2)}$ converges.
Since $\left|\frac{1}{3}\right|<1$, thus $\sum_{r=1}^{\infty}\left(\frac{1}{3}\right)^r$ converges.
Therefore, $\sum_{r=1}^{\infty} \frac{1+\left(\frac{1}{3}\right)^r r !(r+2)}{r !(r+2)}$ converges.$$
\sum_{r=1}^{\infty} \frac{1+\left(\frac{1}{3}\right)^r r !(r+2)}{r !(r+2)}=\frac{1}{2}+\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}+\frac{1}{2}=1$$