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

1. Rewrite the sum using the identity given in the earlier part.

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

2. Write out the first few and last few terms to observe the telescoping cancellation pattern.

$\begin{aligned} \sum_{r=2}^{n} [f(r-1) - f(r)] =& [f(1) - f(2)] \\ &+ [f(2) - f(3)] \\ &+ [f(3) - f(4)] \\ &+ \dots \\ &+ [f(n-1) - f(n)] \end{aligned}$

3. Cancel the intermediate terms and evaluate the remaining terms to find the final expression.

$\begin{aligned} \sum_{r=2}^{n} [f(r-1) - f(r)] &= f(1) - f(n) \\ &= \frac{1}{(1-1)!} - \frac{n}{(n-1)!} \\ &= \frac{1}{0!} - \frac{n}{(n-1)!} \\ &= 1 - \frac{n}{(n-1)!} \end{aligned}$

1. Set up the infinite series as the limit of the partial sum from part (i).

From part (i), we know that $\displaystyle\sum_{r=2}^{n} \frac{r^2-3r+1}{(r-1)!} = 1 - \frac{n}{(n-1)!}$.

To find the sum to infinity, we consider the limit as $n \to \infty$:
$\sum_{r=2}^{\infty} \frac{r^2-3r+1}{(r-1)!} = \lim_{n \to \infty} \left( 1 - \frac{n}{(n-1)!} \right)$

2. Evaluate the limit with justification to find the value.

As $n \to \infty$, the factorial expression $(n-1)!$ in the denominator grows significantly faster than the linear expression $n$ in the numerator.
Therefore, $\dfrac{n}{(n-1)!} \to 0$.

$\lim_{n \to \infty} \left( 1 - \frac{n}{(n-1)!} \right) = 1 - 0 = 1$