1. **Stating the problem:** We want to analyze how the convergence of the infinite series $$\sum_{n=r}^\infty \frac{(n-r)!}{n!}$$ depends on the integer $r$. 2. **Understanding the terms:** The general term of the series is $$a_n = \frac{(n-r)!}{n!}$$ for $n \geq r$. 3. **Simplify the term:** Recall that $$n! = n \times (n-1) \times \cdots \times (n-r+1) \times (n-r)!$$ so we can write $$a_n = \frac{(n-r)!}{n!} = \frac{(n-r)!}{n \times (n-1) \times \cdots \times (n-r+1) \times (n-r)!} = \frac{1}{n \times (n-1) \times \cdots \times (n-r+1)}$$ 4. **Rewrite the denominator:** The denominator is a product of $r$ consecutive integers starting from $n-r+1$ up to $n$: $$a_n = \frac{1}{\prod_{k=0}^{r-1} (n-k)}$$ 5. **Asymptotic behavior:** For large $n$, each factor behaves like $n$, so $$a_n \approx \frac{1}{n^r}$$ 6. **Convergence test:** The series behaves like $$\sum_{n=r}^\infty \frac{1}{n^r}$$ which is a $p$-series with $p = r$. 7. **Known result for $p$-series:** The series $$\sum_{n=1}^\infty \frac{1}{n^p}$$ converges if and only if $p > 1$. 8. **Conclusion:** - If $r > 1$, the series converges. - If $r \leq 1$, the series diverges. **Final answer:** The convergence of the series depends on $r$ such that it converges if and only if $r > 1$.