1. Problem 11: Show that if the sequence $x_n$ of real numbers converges to $x$ then the sequence $s_n=\frac{x_1+\cdots+x_n}{n}$ also converges to $x$. 1.1. Formula and rules: we use the definition of limit of a sequence: $x_n\to x$ means for every $\epsilon>0$ there exists $N$ with $|x_n-x|<\epsilon$ for all $n\ge N$. 1.2. Proof: let $\epsilon>0$. 1.3. Choose $N$ so that $|x_n-x|<\epsilon$ for all $n\ge N$. 1.4. Write $$s_n-x=\frac{1}{n}\sum_{k=1}^n(x_k-x)=\frac{1}{n}\sum_{k=1}^{N-1}(x_k-x)+\frac{1}{n}\sum_{k=N}^n(x_k-x).$$ 1.5. Take absolute values and bound: $$|s_n-x|\le \frac{1}{n}\sum_{k=1}^{N-1}|x_k-x|+\frac{1}{n}\sum_{k=N}^n|x_k-x|.$$ 1.6. For the tail term use $|x_k-x|<\epsilon$ for $k\ge N$ to get $$\frac{1}{n}\sum_{k=N}^n|x_k-x|\le \frac{n-N+1}{n}\epsilon=\Bigl(1-\frac{N-1}{n}\Bigr)\epsilon.$$ 1.7. Note $\lim_{n\to\infty}\Bigl(1-\frac{N-1}{n}\Bigr)=1$ and the finite-first-sum term satisfies $$\frac{1}{n}\sum_{k=1}^{N-1}|x_k-x|\to 0\quad\text{as }n\to\infty.$$ 1.8. Conclude that for large $n$ the first term is $<\epsilon$ and the second term is $\le\epsilon$, so $|s_n-x|<2\epsilon$ eventually, and hence $s_n\to x$. 1.9. (This is the Ces\'aro mean result.) 2. Problem 12: Determine convergence/divergence for each series and give the test. 2.a. $\sum_{n=1}^\infty\frac{1}{2^n}$. 2.a.1. This is a geometric series with ratio $r=\frac{1}{2}$ and first term $\frac{1}{2}$. 2.a.2. Since $|r|<1$ the series converges and sums to $\frac{1/2}{1-1/2}=1$. 2.b. $\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}$. 2.b.1. Use the Alternating Series Test (Leibniz): if $a_n\downarrow0$ then $\sum (-1)^n a_n$ converges. 2.b.2. Here $a_n=1/\sqrt{n}$ is positive, decreasing and $a_n\to0$, so the series converges. 2.b.3. It is conditional (not absolutely) because $\sum 1/\sqrt{n}$ diverges ($p=1/2\le1$). 2.c. $\sum_{n=1}^\infty\frac{1}{n!}$. 2.c.1. Use the Ratio Test: $\displaystyle\frac{a_{n+1}}{a_n}=\frac{1/(n+1)!}{1/n!}=\frac{1}{n+1}\to0<1$. 2.c.2. Hence the series converges absolutely; in fact the sum equals $e-1$. 2.d. $\sum_{n=1}^\infty\frac{\sin n}{n^2}$. 2.d.1. Use absolute comparison: $|\sin n|\le1$ so $\displaystyle\left|\frac{\sin n}{n^2}\right|\le\frac{1}{n^2}$. 2.d.2. Because $\sum 1/n^2$ converges ($p=2>1$), by comparison $\sum \sin n/n^2$ converges absolutely. 2.e. $\sum_{n=1}^\infty\frac{3n^2+2n}{2^n}$. 2.e.1. Use the Ratio Test: let $a_n=\frac{3n^2+2n}{2^n}$ and compute $$\frac{a_{n+1}}{a_n}=\frac{3(n+1)^2+2(n+1)}{3n^2+2n}\cdot\frac{1}{2}.$$ 2.e.2. Divide numerator and denominator by $n^2$ and show cancellation: $$\frac{3(1+1/n)^2+2(1/n+1/n^2)}{3+2/n}\cdot\frac{1}{2}.$$ 2.e.3. As $n\to\infty$ this ratio tends to $\frac{3}{3}\cdot\frac{1}{2}=\frac{1}{2}<1$, so the series converges. 2.f. $\sum_{n=1}^\infty\log\Bigl(\frac{n+1}{n}\Bigr)$. 2.f.1. This telescopes: partial sum $S_N=\sum_{k=1}^N\log\frac{k+1}{k}=\log\prod_{k=1}^N\frac{k+1}{k}=\log(N+1)$. 2.f.2. Since $\log(N+1)\to\infty$ as $N\to\infty$, the series diverges to infinity. 2.f.3. Illustrate cancellation in the product: $(2/1)\cdot(3/2)\cdots((N+1)/N)$ where intermediate factors cancel, e.g. $\frac{\cancel{2}}{\cancel{2}}$. 2.g. $\sum_{n=1}^\infty\frac{1}{2^n+1}$. 2.g.1. Use Limit Comparison Test with $b_n=1/2^n$. Compute $$\lim_{n\to\infty}\frac{1/(2^n+1)}{1/2^n}=\lim_{n\to\infty}\frac{2^n}{2^n+1}=1.$$ 2.g.2. Since $\sum 1/2^n$ converges, so does $\sum 1/(2^n+1)$. 2.h. $\sum_{n=1}^\infty n^4 e^{-n^2}$. 2.h.1. Use the root test: $a_n^{1/n}=(n^4)^{1/n}e^{-n}=n^{4/n}e^{-n}\to e^{-\infty}=0<1$. 2.h.2. Hence the series converges absolutely. 2.i. $\sum_{n=1}^\infty\Bigl(\frac{7n}{8n+1}\Bigr)^n$. 2.i.1. Use the root test: $\lim_{n\to\infty} a_n^{1/n}=\lim_{n\to\infty}\frac{7n}{8n+1}=\frac{7}{8}<1$. 2.i.2. Thus the series converges. 2.j. $\sum_{n=1}^\infty\frac{n+\sqrt{n}}{2n^3-1}$. 2.j.1. Compare to $1/n^2$: divide numerator and denominator by $n^3$ to see $$\frac{n+\sqrt{n}}{2n^3-1}=\frac{1/n^2+1/n^{5/2}}{2-1/n^3}\sim\frac{1}{2n^2}.$$ 2.j.2. Since $\sum 1/n^2$ converges, by comparison this series converges. 2.j.3. When simplifying we divided by $\cancel{n^3}$ in numerator and denominator for clarity. 2.k. $\sum_{n=1}^\infty (-1)^n\frac{n}{100n+100^2}$. 2.k.1. Check the term test: compute $\lim_{n\to\infty}\frac{n}{100n+100^2}=\frac{1}{100}$. 2.k.2. Thus the terms do not tend to $0$, so the series diverges by the divergence (term) test. 2.k.3. When dividing top and bottom by $n$ we show $\frac{\cancel{n}}{\cancel{n}}$ to indicate cancellation. 3. Problem 13: Show $\sum_{n=1}^\infty\frac{1}{n^n}$ converges given $n^n\ge 2^n$ for $n\ge2$. 3.1. For $n\ge2$ we have $0\le\frac{1}{n^n}\le\frac{1}{2^n}$. 3.2. Since $\sum_{n=2}^\infty\frac{1}{2^n}$ converges, by comparison $\sum_{n=2}^\infty\frac{1}{n^n}$ converges, hence the whole series converges. 4. Problem 14: Show $\sum_{n=2}^\infty\frac{1}{n\log n}$ diverges. 4.1. Use the integral test: consider $\int_2^\infty\frac{1}{x\log x}\,dx$. 4.2. Substitute $u=\log x$, $du=dx/x$, so integral equals $\int_{\log 2}^{\infty}\frac{1}{u}\,du=\lim_{T\to\infty}\ln u\Big|_{\log 2}^T=\infty$. 4.3. By the integral test the series diverges. 5. Problem 15: Metrics on $\mathbb{R}^2$. 5.a. $d(x,y)=|x_1-y_1|+|x_2-y_2|$. 5.a.1. This is the $\ell^1$ norm distance; it satisfies nonnegativity, symmetry, and identity of indiscernibles, and the triangle inequality follows from the triangle inequality in $\mathbb{R}$ applied coordinatewise. 5.a.2. Therefore it is a metric. 5.b. $d(x,y)=\sup\{|x_1-y_1|,|x_2-y_2|\}$. 5.b.1. This is the $\ell^\infty$ norm; it also satisfies all metric axioms and the triangle inequality $\|x-z\|_\infty\le\|x-y\|_\infty+\|y-z\|_\infty$. 5.b.2. Hence it is a metric. 6. Problem 16: Metrics on $\mathbb{R}$. 6.a. $d(x,y)=|x-y|^3$. 6.a.1. Nonnegativity and symmetry hold, but the triangle inequality fails in general because $(a+b)^3\not\le a^3+b^3$. 6.a.2. Counterexample: $x=0,y=1,z=2$: $d(0,2)=|0-2|^3=8$ while $d(0,1)+d(1,2)=1+1=2$, so triangle fails; not a metric. 6.b. $d(x,y)=|e^x-e^y|$. 6.b.1. Because $f(x)=e^x$ is injective, $d(x,y)=|f(x)-f(y)|$ inherits metric properties from the absolute value on $\mathbb{R}$, so it is a metric. 6.c. $d(x,y)=|\sin x-\sin y|$. 6.c.1. This fails identity of indiscernibles because $\sin x=\sin y$ can occur with $x\ne y$, so $d(x,y)=0$ while $x\ne y$; not a metric. 6.d. $d(x,y)=(x-y)^2$. 6.d.1. Symmetry and nonnegativity hold but the triangle inequality fails (similar to the $p=2$ power example); not a metric. 6.e. $d(x,y)=\min\{1,|x-y|\}$. 6.e.1. This truncation is a metric: identity holds, symmetry holds, and triangle inequality follows since $\min\{1,a+b\}\le\min\{1,a\}+\min\{1,b\}$. 6.e.2. So it is a metric. 6.f. $d(x,y)=\begin{cases}|x-y|&\text{if }|x-y|\le1\\1&\text{if }|x-y|\ge1\end{cases}$. 6.f.1. This is the same function as in (e) and is a metric for the same reason. 6.g. $d(x,y)=\sqrt{|x-y|}$. 6.g.1. For $00$ choose $\delta=\epsilon/3$ so that if $|x-2|<\delta$ then $|f(x)-f(2)|=3|x-2|<3\delta=\epsilon$. 13.4. When dividing by $3$ we explicitly cancel factors as in $\displaystyle\frac{\cancel{3}|x-2|}{\cancel{3}}$ to indicate the step of obtaining $|x-2|<\epsilon/3$. 13.5. Thus $f$ is continuous at $x=2$ by the $\epsilon$-$\delta$ definition. End of solutions.