Given $f(r)=\dfrac{r}{(r-1)!}$, show that $f(r-1)-f(r)=\dfrac{r^2-3r+1}{(r-1)!}$.

1. Substitute the expression using the identity proven in the main part of the question.

$\sum_{r=2}^n \dfrac{r^2-3r+1}{(r-1)!} = \sum_{r=2}^n [f(r-1) - f(r)]$

2. Expand the sum to observe the cancellation of intermediate terms (method of differences).

$= [f(1) - f(2)] + [f(2) - f(3)] + \dots + [f(n-1) - f(n)]$

3. Cancel the intermediate terms and evaluate the remaining initial and final terms.

$= f(1) - f(n)$
$= \dfrac{1}{(1-1)!} - \dfrac{n}{(n-1)!}$
$= \dfrac{1}{0!} - \dfrac{n}{(n-1)!}$
$= 1 - \dfrac{n}{(n-1)!}$

1. Identify the limit of the sum to $n$ terms as $n$ approaches infinity to justify the convergence

From part (i), the sum to $n$ terms is $S_n = 1 - \frac{n}{(n-1)!}$.

As $n \to \infty$, the factorial term $(n-1)!$ grows significantly faster than the linear term $n$, which means $\frac{n}{(n-1)!} \to 0$.

2. State the value of the sum to infinity

$\therefore \sum_{r=2}^\infty \frac{r^2-3r+1}{(r-1)!} = \lim_{n \to \infty} \left( 1 - \frac{n}{(n-1)!} \right) = 1 - 0 = 1$