The sum of the first $n$ terms of a sequence, $u_r$, is given by $\displaystyle\sum_{r=1}^n u_r = 1 - \dfrac{n}{(n+1)!}$.

1. Apply the relationship between the $n$th term and the sum of the first $n$ terms.

$u_n = S_n - S_{n-1}$
$u_n = \left( 1 - \dfrac{n}{(n+1)!} \right) - \left( 1 - \dfrac{n-1}{n!} \right)$

2. Find a common denominator to combine the terms into a single fraction.

$u_n = \dfrac{n-1}{n!} - \dfrac{n}{(n+1)!}$
$u_n = \dfrac{(n-1)(n+1)}{(n+1)!} - \dfrac{n}{(n+1)!}$

3. Simplify the numerator to express the answer as a single algebraic fraction.

$u_n = \dfrac{n^2 - 1 - n}{(n+1)!}$
$u_n = \dfrac{n^2 - n - 1}{(n+1)!}$

1. Express the required sum as the difference of two sums starting from the first term.

$\sum_{r=5}^n u_r = \sum_{r=1}^n u_r - \sum_{r=1}^4 u_r$

2. Substitute the given sum formula to evaluate and simplify the expression.

$\sum_{r=5}^n u_r = \left( 1 - \frac{n}{(n+1)!} \right) - \left( 1 - \frac{4}{(4+1)!} \right)$
$= 1 - \frac{n}{(n+1)!} - 1 + \frac{4}{120}$
$= \frac{1}{30} - \frac{n}{(n+1)!}$

3. Conclude the proof by showing the subtracted term is always strictly positive.

Since $n \ge 5$, $n > 0$ and $(n+1)! > 0$.
Therefore, $\frac{n}{(n+1)!} > 0$.

Subtracting a strictly positive number from $\frac{1}{30}$ yields a value strictly less than $\frac{1}{30}$.

Thus, $\sum_{r=5}^n u_r < \frac{1}{30}$ for all $n \ge 5$.

1. Evaluate the limit of the sum to $n$ terms as $n$ approaches infinity and state the conclusion.

As $n \to \infty$, $\frac{n}{(n+1)!} \to 0$.

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

Since the limit of the sum to $n$ terms approaches a finite constant ($1$) as $n \to \infty$, the series is convergent.