1. **Problem:** Show that if $(x_n)$ converges to $x$, then the sequence $s_n = \frac{x_1 + x_2 + \cdots + x_n}{n}$ also converges to $x$. 2. **Formula and idea:** Since $x_n \to x$, for every $\epsilon > 0$, there exists $N$ such that for all $n \geq N$, $|x_n - x| < \epsilon$. 3. **Proof:** $$s_n - x = \frac{1}{n} \sum_{k=1}^n (x_k - x) = \frac{1}{n} \sum_{k=1}^N (x_k - x) + \frac{1}{n} \sum_{k=N+1}^n (x_k - x)$$ 4. The first sum is fixed (finite terms), so $\left|\frac{1}{n} \sum_{k=1}^N (x_k - x)\right| \leq \frac{C}{n}$ for some constant $C$. 5. For the second sum, since $|x_k - x| < \epsilon$ for $k > N$, $$\left|\frac{1}{n} \sum_{k=N+1}^n (x_k - x)\right| \leq \frac{n-N}{n} \epsilon \leq \epsilon$$ 6. Combining, $$|s_n - x| \leq \frac{C}{n} + \epsilon$$ 7. As $n \to \infty$, $\frac{C}{n} \to 0$, so $|s_n - x| \to \epsilon$ for any $\epsilon > 0$, hence $s_n \to x$. --- 8. **Problem:** Determine convergence/divergence of series and state test. (a) $\sum_{n=1}^\infty \frac{1}{2^n}$ - Geometric series with ratio $r=\frac{1}{2}$, $|r|<1$, convergent. (b) $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n!}}$ - Terms $\to 0$ rapidly, factorial grows fast. - Absolute convergence by ratio test: $$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac{1/\sqrt{(n+1)!}}{1/\sqrt{n!}} = \lim_{n\to\infty} \frac{\sqrt{n!}}{\sqrt{(n+1)!}} = \lim_{n\to\infty} \frac{1}{\sqrt{n+1}} = 0 < 1$$ - Convergent absolutely. (c) $\sum_{n=1}^\infty \frac{1}{2n-1}$ - Compare to harmonic series $\sum \frac{1}{n}$, diverges. - Divergent by comparison test. (d) $\sum_{n=1}^\infty \frac{\sin n}{n^2}$ - $|\sin n| \leq 1$, so $|a_n| \leq \frac{1}{n^2}$. - $\sum \frac{1}{n^2}$ converges (p-series, $p=2>1$). - Convergent absolutely by comparison. (e) $\sum_{n=1}^\infty \frac{3n^2 + 2n}{2^n}$ - Exponential denominator dominates polynomial numerator. - Ratio test: $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{3(n+1)^2 + 2(n+1)}{2^{n+1}} \cdot \frac{2^n}{3n^2 + 2n} = \frac{1}{2} < 1$$ - Convergent. (f) $\sum_{n=1}^\infty \log\left(\frac{n+1}{n}\right)$ - Telescopes: $$S_N = \sum_{n=1}^N \log\left(\frac{n+1}{n}\right) = \log(N+1)$$ - Diverges to infinity. (g) $\sum_{n=1}^\infty \frac{1}{2^n + 1}$ - Compare to $\sum \frac{1}{2^n}$ convergent geometric. - Convergent by comparison. (h) $\sum_{n=1}^\infty n^4 e^{-n^2}$ - Exponential decay dominates polynomial growth. - Convergent by comparison with $e^{-n}$. (i) $\sum_{n=1}^\infty \left(\frac{n}{8n+1}\right)^n$ - Limit of base: $$\lim_{n\to\infty} \frac{n}{8n+1} = \frac{1}{8} < 1$$ - So terms behave like $(1/8)^n$, convergent by root test. (j) $\sum_{n=1}^\infty \frac{n + \sqrt{n}}{2n^3 - 1}$ - Dominant term numerator $n$, denominator $2n^3$. - Terms behave like $\frac{n}{2n^3} = \frac{1}{2n^2}$. - Convergent by comparison with $1/n^2$. (k) $\sum_{n=1}^\infty (-1)^n \frac{n}{100n + 100}$ - Terms $a_n = \frac{n}{100n + 100} \to \frac{1}{100} \neq 0$. - Divergent by nth term test. --- 10. **Problem:** Show $\sum_{n=1}^\infty \frac{1}{n^n}$ converges given $n^n \geq 2^n$ for $n \geq 2$. 11. **Proof:** Since $n^n \geq 2^n$, $$\frac{1}{n^n} \leq \frac{1}{2^n}$$ 12. $\sum \frac{1}{2^n}$ converges (geometric series), so by comparison test $\sum \frac{1}{n^n}$ converges. --- 13. **Problem:** Show $\sum_{n=2}^\infty \frac{1}{n \log n}$ diverges. 14. **Proof:** Integral test: $$\int_2^\infty \frac{1}{x \log x} dx = \lim_{t \to \infty} \log(\log t) - \log(\log 2) = \infty$$ 15. So series diverges. --- 16. **Problem:** Decide if $d$ is a metric on $\mathbb{R}^2$. (a) $d(x,y) = |x_1 - y_1| + |x_2 - y_2|$ - Satisfies positivity, symmetry, triangle inequality. - Metric. (b) $d(x,y) = \sup\{|x_1 - y_1|, |x_2 - y_2|\}$ - Also satisfies metric properties. - Metric. --- 17. **Problem:** Decide if $d$ is a metric on $\mathbb{R}$. (a) $d(x,y) = |x - y|^3$ - Positivity and symmetry hold. - Triangle inequality fails because $|a+b|^3 \leq |a|^3 + |b|^3$ is false. - Not a metric. (b) $d(x,y) = |e^x - e^y|$ - $e^x$ is injective and continuous. - Triangle inequality holds by properties of absolute value. - Metric. (c) $d(x,y) = |\sin x - \sin y|$ - $\sin$ is not injective. - $d(x,y) = 0$ may hold for $x \neq y$. - Not a metric. (d) $d(x,y) = (x - y)^2$ - Triangle inequality fails. - Not a metric. (e) $d(x,y) = \min\{1, |x - y|\}$ - Satisfies all metric properties. - Metric. (f) $d(x,y) = \begin{cases} |x - y| & |x - y| \leq 1 \\ 1 & |x - y| > 1 \end{cases}$ - Triangle inequality holds. - Metric. (g) $d(x,y) = \sqrt{|x - y|}$ - Triangle inequality holds (proved later). - Metric. (h) $d'(x,y) = \min\{1, d(x,y)\}$ with $d$ metric - Metric. (i) $d(x,y) = \begin{cases} 1 & x \neq y \\ 0 & x = y \end{cases}$ - Discrete metric. - Metric. --- 18. **Problem:** Is $d(x,y) = |1/x - 1/y|$ a metric on $\mathbb{R} \setminus \{0\}$? - Positivity, symmetry hold. - Triangle inequality holds by absolute value. - Metric. --- 19. **Problem:** Show $d'(x,y) = \frac{d(x,y)}{1 + d(x,y)}$ is a metric if $d$ is metric. - Positivity and symmetry clear. - Triangle inequality: $$d'(x,z) = \frac{d(x,z)}{1 + d(x,z)} \leq \frac{d(x,y) + d(y,z)}{1 + d(x,y) + d(y,z)} \leq \frac{d(x,y)}{1 + d(x,y)} + \frac{d(y,z)}{1 + d(y,z)} = d'(x,y) + d'(y,z)$$ - So $d'$ is metric. --- 20. **Problem:** Show $d(x,y) = |f(x) - f(y)|$ is metric if $f$ bijection on $\mathbb{R}$. - Positivity: $d(x,y) = 0 \Rightarrow f(x) = f(y) \Rightarrow x = y$ since $f$ injective. - Symmetry and triangle inequality hold by absolute value. - Metric. --- 21. **Problem:** Show $d(x,y) = \sqrt{|x - y|}$ is metric. - Positivity and symmetry clear. - Triangle inequality: Use concavity of $\sqrt{\cdot}$ and Minkowski inequality: $$\sqrt{|x-z|} \leq \sqrt{|x-y|} + \sqrt{|y-z|}$$ - So metric. --- 22. **Problem:** 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 everywhere. (d) $\frac{1}{1-x}$ continuous on intervals excluding $x=1$, not uniformly continuous on intervals containing $1$. (e) $\sqrt{x} - 1$ continuous on $[0,1]$, uniformly continuous on $[0,1]$, not uniformly continuous on $(2,\infty)$. --- 23. **Problem:** Show $f(x) = \frac{x^3}{1 + x^2}$ continuous on $\mathbb{R}$. - $f$ is quotient of polynomials with denominator never zero. - Continuous everywhere. - Not uniformly continuous on $\mathbb{R}$ because derivative unbounded. --- 24. **Problem:** Prove $f(x) = 3x - 5$ continuous at $x=2$ using definition. - For $\epsilon > 0$, choose $\delta = \frac{\epsilon}{3}$. - If $|x - 2| < \delta$, then $$|f(x) - f(2)| = |3x - 5 - (6 - 5)| = |3x - 6| = 3|x - 2| < 3 \delta = \epsilon$$ - Hence continuous at $x=2$.