1. **State the problem:** We want to prove using the definition of convergence that $$\lim_{n \to \infty} \frac{5}{1+n^2} = 0.$$\n\n2. **Recall the definition of convergence:** A sequence $a_n$ converges to a limit $L$ if for every $\epsilon > 0$, there exists an integer $N$ such that for all $n > N$, $$|a_n - L| < \epsilon.$$\n\n3. **Apply the definition to our sequence:** Here, $a_n = \frac{5}{1+n^2}$ and $L = 0$. We want to show that for every $\epsilon > 0$, there exists $N$ such that for all $n > N$, $$\left| \frac{5}{1+n^2} - 0 \right| < \epsilon.$$\n\n4. **Simplify the inequality:** Since the expression is positive, $$\frac{5}{1+n^2} < \epsilon.$$\n\n5. **Solve for $n$:** Multiply both sides by $1+n^2$ and divide by $\epsilon$ (assuming $\epsilon > 0$), we get $$5 < \epsilon (1+n^2) \implies \frac{5}{\epsilon} < 1 + n^2 \implies n^2 > \frac{5}{\epsilon} - 1.$$\n\n6. **Choose $N$:** Let $$N = \left\lceil \sqrt{\frac{5}{\epsilon} - 1} \right\rceil.$$ Then for all $n > N$, $$\frac{5}{1+n^2} < \epsilon.$$\n\n7. **Conclusion:** By the definition of convergence, the sequence $\frac{5}{1+n^2}$ converges to 0 as $n$ approaches infinity.