1. **Problem statement:** Given the sequence defined by $a_1 = 3$ and the recurrence relation $$a_{n+1} = \frac{2a_n + 5}{a_n + 1},$$ determine whether the sequence $(a_n)$ converges and if so, find its limit. 2. **Formula and approach:** To find the limit of a convergent sequence defined recursively, assume the sequence converges to a limit $L$. Then, as $n \to \infty$, both $a_n$ and $a_{n+1}$ approach $L$. So we set: $$L = \frac{2L + 5}{L + 1}$$ 3. **Solve for $L$:** Multiply both sides by $L + 1$ to clear the denominator: $$L(L + 1) = 2L + 5$$ $$L^2 + L = 2L + 5$$ Bring all terms to one side: $$L^2 + L - 2L - 5 = 0$$ $$L^2 - L - 5 = 0$$ 4. **Use the quadratic formula:** For $L^2 - L - 5 = 0$, the solutions are: $$L = \frac{1 \pm \sqrt{1^2 - 4 \times 1 \times (-5)}}{2} = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$$ 5. **Evaluate the roots:** $$L_1 = \frac{1 + \sqrt{21}}{2} \approx 3.79$$ $$L_2 = \frac{1 - \sqrt{21}}{2} \approx -2.79$$ 6. **Check stability and initial value:** Since $a_1 = 3$ is closer to $L_1 \approx 3.79$ and the sequence is defined by a rational function, the sequence will converge to the stable fixed point near the initial value. Testing a few terms confirms convergence to $L_1$. **Final answer:** The sequence $(a_n)$ converges and its limit is $$\boxed{\frac{1 + \sqrt{21}}{2}}.$$