1. **Stating the problem:** Find the limit $$\lim_{n \to \infty} \frac{1 + r + r^2 + \cdots + r^n}{n + r} - \frac{n}{r}$$ where $r$ is a constant. 2. **Recall the formula for the sum of a geometric series:** $$\sum_{k=0}^n r^k = \frac{1 - r^{n+1}}{1 - r} \quad \text{for } r \neq 1$$ 3. **Rewrite the numerator using the geometric series sum:** $$1 + r + r^2 + \cdots + r^n = \frac{1 - r^{n+1}}{1 - r}$$ 4. **Substitute into the limit expression:** $$\lim_{n \to \infty} \frac{\frac{1 - r^{n+1}}{1 - r}}{n + r} - \frac{n}{r} = \lim_{n \to \infty} \frac{1 - r^{n+1}}{(1 - r)(n + r)} - \frac{n}{r}$$ 5. **Analyze the behavior as $n \to \infty$ depending on $|r|$:** - If $|r| < 1$, then $r^{n+1} \to 0$. - If $|r| > 1$, then $r^{n+1} \to \infty$ or diverges. - If $r = 1$, the sum is $n+1$. 6. **Case 1: $|r| < 1$** $$\lim_{n \to \infty} \frac{1 - 0}{(1 - r)(n + r)} - \frac{n}{r} = \lim_{n \to \infty} \frac{1}{(1 - r)(n + r)} - \frac{n}{r}$$ The first term tends to 0 as $n \to \infty$, so the limit behaves like $$- \frac{n}{r} \to -\infty$$ 7. **Case 2: $r = 1$** Sum is $n+1$, so $$\lim_{n \to \infty} \frac{n+1}{n+1} - \frac{n}{1} = \lim_{n \to \infty} 1 - n = -\infty$$ 8. **Case 3: $|r| > 1$** Since $r^{n+1}$ dominates, $$\frac{1 - r^{n+1}}{(1 - r)(n + r)} \approx \frac{-r^{n+1}}{(1 - r)(n + r)}$$ As $n \to \infty$, numerator grows exponentially, denominator grows linearly, so the fraction tends to $\pm \infty$ depending on $r$. 9. **Conclusion:** - For $|r| \leq 1$, the limit diverges to $-\infty$. - For $|r| > 1$, the limit diverges to infinity or negative infinity depending on the sign of $r^{n+1}$. **Final answer:** The limit does not converge to a finite value for any $r$.