1. Problem 11: Show that if $(x_n)$ converges to $x$, then the sequence $(s_n)$ where $s_n = \frac{x_1 + x_2 + \cdots + x_n}{n}$ also converges to $x$. 2. Recall the definition of convergence: A sequence $(x_n)$ converges to $x$ if for every $\epsilon > 0$, there exists $N$ such that for all $n \geq N$, $|x_n - x| < \epsilon$. 3. We want to show $\lim_{n \to \infty} s_n = x$. Note that $$s_n - x = \frac{1}{n} \sum_{k=1}^n (x_k - x).$$ 4. For any $\epsilon > 0$, since $x_n \to x$, there exists $N_1$ such that for all $k \geq N_1$, $|x_k - x| < \epsilon$. 5. Split the sum: $$s_n - x = \frac{1}{n} \left( \sum_{k=1}^{N_1 - 1} (x_k - x) + \sum_{k=N_1}^n (x_k - x) \right).$$ 6. The first sum is fixed (finite terms), call it $M = \sum_{k=1}^{N_1 - 1} (x_k - x)$. 7. Then $$|s_n - x| \leq \frac{|M|}{n} + \frac{1}{n} \sum_{k=N_1}^n |x_k - x| < \frac{|M|}{n} + \frac{n - N_1 + 1}{n} \epsilon \leq \frac{|M|}{n} + \epsilon.$$ 8. As $n \to \infty$, $\frac{|M|}{n} \to 0$, so for large $n$, $|s_n - x| < \epsilon + \delta$ for any small $\delta$. 9. Hence $s_n \to x$. --- 10. Problem 12: Determine convergence/divergence of series and state test used. (a) $\sum_{n=1}^\infty \frac{1}{2^n}$ is a geometric series with ratio $\frac{1}{2} < 1$, so convergent by geometric series test. (b) $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n!}}$: Since $\sqrt{n!}$ grows faster than any exponential, terms go to zero rapidly. By absolute convergence test (ratio test), convergent. (c) $\sum_{n=1}^\infty \frac{1}{2^n n!}$: Terms positive and decrease rapidly, ratio test shows convergence. (d) $\sum_{n=1}^\infty \frac{\sin n}{n^2}$: Since $|\sin n| \leq 1$, by comparison with $\sum \frac{1}{n^2}$ (p-series, $p=2>1$), convergent absolutely. (e) $\sum_{n=1}^\infty \frac{3n^2 + 2n}{2^n}$: Numerator polynomial, denominator exponential, ratio test shows convergence. (f) $\sum_{n=1}^\infty \log\left(\frac{n+1}{n}\right) = \sum \log\left(1 + \frac{1}{n}\right)$: This telescopes to $\log(n+1)$ which diverges to infinity, so series diverges. (g) $\sum_{n=1}^\infty \frac{1}{2^n + 1}$: Compare with $\sum \frac{1}{2^n}$, convergent by comparison test. (h) $\sum_{n=1}^\infty n^4 e^{-n^2}$: Exponential decay dominates polynomial growth, convergent by comparison. (i) $\sum_{n=1}^\infty \left(\frac{n}{8n+1}\right)^n$: Since $\frac{n}{8n+1} \to \frac{1}{8} < 1$, terms behave like $(\frac{1}{8})^n$, convergent by root test. (j) $\sum_{n=1}^\infty \frac{n + \sqrt{n}}{2n^3 - 1}$: Dominated by $\frac{n}{2n^3} = \frac{1}{2n^2}$, convergent by comparison. (k) $\sum_{n=1}^\infty (-1)^n \frac{n}{100n + 100}$: Terms $\to \frac{1}{100}$ not zero, so series diverges by nth term test. --- 11. Problem 13: Given $n^n \geq 2^n$ for $n \geq 2$, show $\sum_{n=1}^\infty \frac{1}{n^n}$ converges. 12. Since $n^n \geq 2^n$, then $\frac{1}{n^n} \leq \frac{1}{2^n}$ for $n \geq 2$. 13. $\sum_{n=2}^\infty \frac{1}{2^n}$ converges (geometric series), so by comparison test, $\sum_{n=2}^\infty \frac{1}{n^n}$ converges. 14. Adding finite term $n=1$ does not affect convergence. --- 15. Problem 14: Show $\sum_{n=2}^\infty \frac{1}{n \log n}$ diverges. 16. Use integral test: Consider $$\int_2^\infty \frac{1}{x \log x} dx.$$ 17. Substitute $t = \log x$, $dt = \frac{1}{x} dx$, integral becomes $$\int_{\log 2}^\infty \frac{1}{t} dt,$$ which diverges (harmonic integral). 18. Hence series diverges. --- 19. Problem 15: Decide if given functions define metrics on $\mathbb{R}^2$. (a) $d(x,y) = |x_1 - y_1| + |x_2 - y_2|$ is the Manhattan metric, satisfies all metric properties. (b) $d(x,y) = \sup\{|x_1 - y_1|, |x_2 - y_2|\}$ is the sup metric, also a metric. --- 20. Problem 16: Decide if given functions define metrics on $\mathbb{R}$. (a) $d(x,y) = |x - y|^3$: fails triangle inequality, not a metric. (b) $d(x,y) = |e^x - e^y|$: since $e^x$ is injective and metric on image, this is a metric. (c) $d(x,y) = |\sin x - \sin y|$: since $\sin$ is bounded and not injective, triangle inequality fails, not a metric. (d) $d(x,y) = (x - y)^2$: fails triangle inequality, not a metric. (e) $d(x,y) = \min\{1, |x - y|\}$: truncation preserves metric properties, is a metric. (f) $d(x,y) = \begin{cases} |x - y| & |x - y| \leq 1 \\ 1 & |x - y| > 1 \end{cases}$: also a metric. (g) $d(x,y) = \sqrt{|x - y|}$: satisfies metric properties, is a metric. (h) $d'(x,y) = \min\{1, d(x,y)\}$ where $d$ is a metric: also a metric. (i) $d(x,y) = \begin{cases} 1 & x \neq y \\ 0 & x = y \end{cases}$: discrete metric, is a metric. --- 21. Problem 17: $d(x,y) = |1/x - 1/y|$ on $\mathbb{R} \setminus \{0\}$. Check metric properties: - Non-negativity and symmetry hold. - Identity: $d(x,y) = 0 \iff 1/x = 1/y \iff x = y$. - Triangle inequality holds since absolute value is metric on $\mathbb{R}$. Hence $d$ is a metric. --- 22. Problem 18: Given metric $d$, define $d'(x,y) = \frac{d(x,y)}{1 + d(x,y)}$. Show $d'$ is a metric: - Non-negativity and identity: clear. - Symmetry: follows from $d$. - Triangle inequality: Use that $f(t) = \frac{t}{1+t}$ is increasing and subadditive on $[0,\infty)$, so $$d'(x,z) = \frac{d(x,z)}{1 + d(x,z)} \leq \frac{d(x,y) + d(y,z)}{1 + d(x,y) + d(y,z)} \leq \frac{d(x,y)}{1 + d(x,y)} + \frac{d(y,z)}{1 + d(y,z)} = d'(x,y) + d'(y,z).$$ --- 23. Problem 19: For bijection $f: \mathbb{R} \to \mathbb{R}$, define $d(x,y) = |f(x) - f(y)|$. Since $f$ is bijection, $d$ satisfies: - Non-negativity and identity. - Symmetry. - Triangle inequality from absolute value. Hence $d$ is a metric. --- 24. Problem 20: Show $d(x,y) = \sqrt{|x - y|}$ is a metric. Check: - Non-negativity, identity, symmetry: clear. - Triangle inequality: Use concavity of $\sqrt{t}$ and Minkowski inequality to show $$\sqrt{|x - z|} \leq \sqrt{|x - y|} + \sqrt{|y - z|}.$$ Hence $d$ is a metric. --- 25. Problem 21: Continuity and uniform continuity of functions on intervals. (a) $\sin x$: continuous and uniformly continuous on all intervals. (b) $e^x$: continuous everywhere, uniformly continuous on bounded intervals like $[0,1]$, not on $(2, \infty)$. (c) $|x - 1/2| + |x - 3|$: continuous and uniformly continuous on all intervals. (d) $\frac{1}{1 - x}$: continuous on intervals excluding $x=1$, not uniformly continuous on intervals containing points near 1. (e) $\sqrt{x} - 1$: continuous on $[0,1]$ and $(0,1)$, uniformly continuous on $[0,1]$, not uniformly continuous on $(2, \infty)$. --- 26. Problem 22: $f(x) = \frac{x^3}{1 + x^2}$. Since numerator and denominator are polynomials and denominator never zero, $f$ is continuous on $\mathbb{R}$. Not uniformly continuous on $\mathbb{R}$ because derivative grows without bound. --- 27. Problem 23: Prove $f(x) = 3x - 5$ is continuous at $x=2$ using definition. For $\epsilon > 0$, choose $\delta = \frac{\epsilon}{3}$. If $|x - 2| < \delta$, then $$|f(x) - f(2)| = |3x - 5 - (6 - 5)| = |3x - 6| = 3|x - 2| < 3 \delta = \epsilon.$$