1. **State the problem:** We want to analyze the convergence of the infinite series $$\sum_{n=r}^\infty \frac{(n-r)!}{n!}$$ where $r$ is an integer. 2. **Rewrite the general term:** Recall that $$n! = n \times (n-1) \times \cdots \times (n-r+1) \times (n-r)!$$ so we can write $$\frac{(n-r)!}{n!} = \frac{(n-r)!}{n \times (n-1) \times \cdots \times (n-r+1) \times (n-r)!} = \frac{1}{n \times (n-1) \times \cdots \times (n-r+1)}.$$ 3. **Simplify the term:** The denominator is a product of $r$ consecutive integers starting from $n-r+1$ up to $n$. Thus, $$a_n = \frac{1}{n (n-1) \cdots (n-r+1)}.$$ 4. **Analyze convergence:** For large $n$, each factor in the denominator behaves like $n$, so the denominator behaves like $n^r$. Hence, $$a_n \sim \frac{1}{n^r} \text{ as } n \to \infty.$$ 5. **Use the p-series test:** The series $$\sum_{n=r}^\infty \frac{1}{n^r}$$ converges if and only if the exponent $r > 1$. 6. **Conclusion:** - If $r > 1$, the original series converges. - If $r \leq 1$, the series diverges. 7. **Summary:** The convergence depends on the integer $r$ such that the series converges if and only if $r$ is an integer greater than 1.