1. Verify convergence of sequences using definition: (a) Show $\lim_{n \to \infty} \frac{2n+1}{5n+4} = \frac{2}{5}$. Step 1: State the problem: Prove $\forall \epsilon > 0, \exists N$ such that $n > N \Rightarrow \left| \frac{2n+1}{5n+4} - \frac{2}{5} \right| < \epsilon$. Step 2: Simplify the difference: $$\left| \frac{2n+1}{5n+4} - \frac{2}{5} \right| = \left| \frac{(2n+1)5 - 2(5n+4)}{5(5n+4)} \right| = \left| \frac{10n + 5 - 10n - 8}{5(5n+4)} \right| = \left| \frac{-3}{5(5n+4)} \right| = \frac{3}{5(5n+4)}.$$ Step 3: To make this less than $\epsilon$, require: $$\frac{3}{5(5n+4)} < \epsilon \Rightarrow 5(5n+4) > \frac{3}{\epsilon} \Rightarrow 25n + 20 > \frac{3}{\epsilon} \Rightarrow n > \frac{\frac{3}{\epsilon} - 20}{25}.$$ Step 4: Choose $N = \max\left(1, \left\lceil \frac{\frac{3}{\epsilon} - 20}{25} \right\rceil \right)$. Step 5: Then for all $n > N$, the inequality holds, proving convergence. (b) Show $\lim_{n \to \infty} \frac{2n^2}{n^3 + 3} = 0$. Step 1: State problem: $\forall \epsilon > 0, \exists N$ such that $n > N \Rightarrow \left| \frac{2n^2}{n^3 + 3} - 0 \right| < \epsilon$. Step 2: Simplify: $$\left| \frac{2n^2}{n^3 + 3} \right| = \frac{2n^2}{n^3 + 3} < \frac{2n^2}{n^3} = \frac{2}{n}.$$ Step 3: Require $\frac{2}{n} < \epsilon \Rightarrow n > \frac{2}{\epsilon}$. Step 4: Choose $N = \max(1, \lceil \frac{2}{\epsilon} \rceil)$. Step 5: For $n > N$, the inequality holds, so sequence converges to 0. (c) Show $\lim_{n \to \infty} \frac{\sin(n^2)}{\sqrt{n}} = 0$. Step 1: Since $|\sin(n^2)| \leq 1$, $$\left| \frac{\sin(n^2)}{\sqrt{n}} \right| \leq \frac{1}{\sqrt{n}}.$$ Step 2: For $\epsilon > 0$, require $\frac{1}{\sqrt{n}} < \epsilon \Rightarrow n > \frac{1}{\epsilon^2}$. Step 3: Choose $N = \max(1, \lceil \frac{1}{\epsilon^2} \rceil)$. Step 4: For $n > N$, inequality holds, so sequence converges to 0. 2. Prove sequence $x_1=3$, $x_{n+1} = \frac{1}{4 - x_n}$ converges. Step 1: Assume limit $L$ exists, then $$L = \frac{1}{4 - L} \Rightarrow L(4 - L) = 1 \Rightarrow 4L - L^2 = 1 \Rightarrow L^2 - 4L + 1 = 0.$$ Step 2: Solve quadratic: $$L = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}.$$ Step 3: Check which root is valid by iteration and boundedness. Step 4: Show sequence is bounded and monotone to apply Monotone Convergence Theorem. 3. Show $x_1 = \sqrt{2}$, $x_{n+1} = \sqrt{2} + x_n$ converges and find limit. Step 1: This is an arithmetic sequence with constant addition $\sqrt{2}$. Step 2: Since $x_{n+1} = x_n + \sqrt{2}$, sequence diverges to infinity, so it does not converge. 4. Sequence $x_1 = \sqrt{2}$, $x_{n+1} = \sqrt{2 x_n}$ convergence? Step 1: Assume limit $L$ exists. Step 2: Then $$L = \sqrt{2L} \Rightarrow L^2 = 2L \Rightarrow L^2 - 2L = 0 \Rightarrow L(L - 2) = 0.$$ Step 3: So $L = 0$ or $L = 2$. Step 4: Since $x_1 = \sqrt{2} > 0$, and sequence terms are positive, check which fixed point is stable. Step 5: By iteration, sequence converges to $2$. 5. Show $\lim_{n \to \infty} \frac{n^2 + 1}{n^2 + n} = 1$. Step 1: Divide numerator and denominator by $n^2$: $$\frac{1 + \frac{1}{n^2}}{1 + \frac{1}{n}}.$$ Step 2: As $n \to \infty$, $\frac{1}{n^2} \to 0$, $\frac{1}{n} \to 0$, so limit is $\frac{1+0}{1+0} = 1$. 6. $x_n = (-1)^n$, $y_n = (-1)^{n+1}$, is $x_n + y_n$ convergent? Step 1: $x_n + y_n = (-1)^n + (-1)^{n+1} = (-1)^n - (-1)^n = 0$. Step 2: Sequence is constantly zero, so converges to 0. 7. Prove $\lim_{n \to \infty} \frac{2^n}{n!} = 0$. Step 1: For large $n$, factorial grows faster than exponential. Step 2: Use ratio test for sequence terms: $$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}/(n+1)!}{2^n/n!} = \frac{2}{n+1} \to 0 < 1.$$ Step 3: So sequence converges to 0. 8. $x_n = \frac{1}{\ln n}$ for $n \geq 2$, prove $x_n \to 0$ with $\epsilon$-$N$ proof. Step 1: For $\epsilon > 0$, require $$\left| \frac{1}{\ln n} - 0 \right| = \frac{1}{\ln n} < \epsilon \Rightarrow \ln n > \frac{1}{\epsilon} \Rightarrow n > e^{\frac{1}{\epsilon}}.$$ Step 2: Choose $N = \max(2, \lceil e^{\frac{1}{\epsilon}} \rceil)$. Step 3: For $n > N$, inequality holds, so $x_n \to 0$. 9. Show $x_n = -n + \sqrt{n^2 + 3n}$ converges and find limit. Step 1: Rationalize: $$x_n = -n + \sqrt{n^2 + 3n} = \frac{(-n + \sqrt{n^2 + 3n})(-n - \sqrt{n^2 + 3n})}{-n - \sqrt{n^2 + 3n}} = \frac{-n^2 + (n^2 + 3n)}{-n - \sqrt{n^2 + 3n}} = \frac{3n}{-n - \sqrt{n^2 + 3n}}.$$ Step 2: Simplify denominator: $$-n - \sqrt{n^2 + 3n} = -n - n\sqrt{1 + \frac{3}{n}} = -n(1 + \sqrt{1 + \frac{3}{n}}).$$ Step 3: So $$x_n = \frac{3n}{-n(1 + \sqrt{1 + \frac{3}{n}})} = \frac{3}{-(1 + \sqrt{1 + \frac{3}{n}})} = -\frac{3}{1 + \sqrt{1 + \frac{3}{n}}}.$$ Step 4: As $n \to \infty$, $\frac{3}{n} \to 0$, so $$x_n \to -\frac{3}{1 + \sqrt{1 + 0}} = -\frac{3}{1 + 1} = -\frac{3}{2}.$$ 10. Determine limits: (a) $x_n = \frac{n^2 + n + 1}{3n^2 + 4}$. Divide numerator and denominator by $n^2$: $$\frac{1 + \frac{1}{n} + \frac{1}{n^2}}{3 + \frac{4}{n^2}} \to \frac{1 + 0 + 0}{3 + 0} = \frac{1}{3}.$$ (b) $x_n = \frac{\sin(n^2 + 1)}{n^2 + 1}$. Since $|\sin| \leq 1$, $$|x_n| \leq \frac{1}{n^2 + 1} \to 0.$$ (c) $x_n = \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}}$. Rationalize numerator and denominator: Numerator: $$\sqrt{n+2} - \sqrt{n+1} = \frac{(n+2) - (n+1)}{\sqrt{n+2} + \sqrt{n+1}} = \frac{1}{\sqrt{n+2} + \sqrt{n+1}}.$$ Denominator: $$\sqrt{n+1} - \sqrt{n} = \frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}.$$ So $$x_n = \frac{\frac{1}{\sqrt{n+2} + \sqrt{n+1}}}{\frac{1}{\sqrt{n+1} + \sqrt{n}}} = \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+2} + \sqrt{n+1}}.$$ As $n \to \infty$, numerator and denominator behave like $2\sqrt{n}$, so limit is $$\frac{2\sqrt{n}}{2\sqrt{n}} = 1.$$ (d) $x_n = \sin n^2 + \sqrt{n+1} - \sqrt{n}$. Since $|\sin n^2| \leq 1$, and $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \to 0,$$ sequence behaves like $\sin n^2$ oscillating between -1 and 1, so does not converge. 11. If $x_n \to x$, show $s_n = \frac{x_1 + x_2 + \cdots + x_n}{n} \to x$. Step 1: By Cesàro mean theorem, the average of convergent sequence converges to the same limit. Step 2: For $\epsilon > 0$, choose $N$ such that $|x_n - x| < \epsilon$ for $n > N$. Step 3: Then split sum and bound difference to show $|s_n - x| < \epsilon$ for large $n$. 12. Series convergence tests: (a) Series $\sum \frac{1}{n^2}$ converges by p-series test with $p=2 > 1$. (b) Series $\sum \frac{1}{n}$ diverges by p-series test with $p=1$. (c) Series $\sum (-1)^n \frac{1}{n}$ converges by Alternating Series Test. (d) Series $\sum \frac{2^n}{n!}$ converges by Ratio Test: $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{2}{n+1} = 0 < 1.$$