## Sequences in Metric Spaces :::{.theorem title="Bolzano-Weierstrass"} Every bounded sequence has a convergent subsequence. ::: :::{.theorem title="Heine-Borel"} In $\RR^n, X$ is compact $\iff X$ is closed and bounded. ::: :::{.remark} Necessity of $\RR^n$: $X = (\ZZ, d(x,y) = 1)$ is closed, complete, bounded, but not compact since $\theset{1,2,\cdots}$ has no convergent subsequence ::: :::{.proposition title="Converse of Heine-Borel"} Converse holds iff bounded is replaced with totally bounded ::: # Sequences Notation: $\theset{a_n}_{n\in\NN}$ is a **sequence**, $\displaystyle\sum_{i\in\NN} a_i$ is a **series**. ## Known Examples - Known sequences: let \(c\) be a constant. \[ c, c^2, c^3, \ldots &= \theset{c^n}_{n=1}^\infty \to 0 && \forall \abs{c} < 1 \\ \\ \frac{1}{c},\frac{1}{c^2},\frac{1}{c^3},\ldots &= \theset{\frac{1}{c^n}}_{n=1}^\infty \to 0 &&\forall \abs{c} > 1 \\ \\ 1,\frac{1}{2^c},\frac{1}{3^c},\ldots &= \theset{\frac{1}{n^c}}_{n=1}^\infty \to 0 && \forall c > 0 \] ## Convergence :::{.definition title="Convergence of a Sequence"} A sequence $\ts{x_j}$ **converges** to $L$ iff \[ \forall \varepsilon > 0,\, \exists N > 0 \text{ such that } \quad n\geq N \implies \abs{x_n - L} < \eps .\] ::: :::{.theorem title="Squeeze Theorem"} \[ b_n \leq a_n \leq c_n \text{ and } b_n,c_n \to L \implies a_n \to L \] ::: :::{.theorem title="Monotone Convergence Theorem for Sequences"} If $\theset{a_j}$ monotone and bounded, then $a_j \to L = \lim\sup a_i < \infty$. ::: :::{.theorem title="Cauchy Criteria"} $\abs{a_m - a_n} \to 0 \in \RR \implies \theset{a_i}$ converges. ::: ### Checklist - Is the sequence bounded? - $\theset{a_i}$ not bounded $\implies$ not convergent - If bounded, is it monotone? - $\theset{a_i}$ bounded and monotone $\implies$ convergent - Use algebraic properties of limits - Epsilon-delta definition - Algebraic properties and manipulation: - Limits commute with $\pm, \times, \div$ and $\lim C = C$ for constants. - E.g. Divide all terms by \(n\) before taking limit - Clear denominators # Sums ("Series") :::{.definition title="Series"} A **series** is an function of the form \[ f(x) = \sum_{j=1}^\infty c_j x^j .\] ::: ## Known Examples ### Conditionally Convergent \[ \sum_{k=1}^\infty k^p &< \infty &&\iff p \leq 1 \\ \sum_{k=1}^\infty \frac{1}{k^p} &< \infty &&\iff p > 1 \\ \sum_{k=1}^\infty \frac{1}{k} &= \infty && \] ### Convergent \[ \sum_{n=1}^\infty \frac{1}{n^2} & < \infty \\ \sum_{n=1}^\infty \frac{1}{n^3} & < \infty \\ \sum_{n=1}^\infty \frac{1}{n^\frac{3}{2}} & < \infty \\ \sum_{n=1}^\infty \frac{1}{n!} & = e \\ \sum_{n=1}^\infty \frac{1}{c^n} & = \frac{c}{c-1} \\ \sum_{n=1}^\infty (-1)^n \frac{1}{c^n} & = \frac{c}{c+1} \\ \sum_{n=1}^\infty (-1)^n \frac{1}{n} & = \ln 2 \] ### Divergent \[ \sum_{n=1}^\infty \frac{1}{n} = \infty \\ \sum_{n=1}^\infty \frac{1}{\sqrt n} = \infty \] ## Convergence > Useful reference: :::{.definition title="Absolutely Convergent"} \todo[inline]{todo} ::: :::{.remark} $a_n\to 0$ does not imply $\sum a_n < \infty$. Counterexample: the harmonic series. ::: :::{.proposition title="?"} Absolute convergence $\implies$ convergence ::: :::{.proposition title="The Cauchy Criterion"} \[ \limsup a_i \to 0 \implies \sum a_i \text{ converges } \] ::: ### The Big Tests :::{.theorem title="Comparison Test"} \envlist - $a_n < b_n \and \sum b_n < \infty \implies \sum a_n < \infty$ - $b_n < a_n \and \sum b_n = \infty \implies \sum a_n = \infty$ ::: :::{.theorem title="Ratio Test"} \[ R =\lim_{n\to\infty} \abs{\frac{a_{n+1}}{a_n}} \] - \(R < 1\): absolutely convergent - \(R > 1\): divergent - \(R = 1\): inconclusive ::: :::{.theorem title="Root Test"} \[ R = \limsup_{n \to \infty} \sqrt[n]{\abs{a_n}} \] - \(R < 1\): convergent - \(R > 1\): divergent - \(R = 1\): inconclusive ::: :::{.theorem title="Integral Test"} \[ f(n) = a_n \implies \sum a_n < \infty \iff \int_1^\infty f(x) dx < \infty \] ::: :::{.theorem title="Limit Test"} \[ \lim_{n\to\infty}\frac{a_n}{b_n} = L < \infty \implies \sum a_n < \infty \iff \sum b_n < \infty \] ::: :::{.theorem title="Alternating Series Test"} \[ a_n \downarrow 0 \implies \sum (-1)^n a_n < \infty \] ::: :::{.theorem title="Weierstrass $M\dash$Test"} \[ \sum_{n=1}^\infty \norm{f_n}_\infty < \infty \implies \exists f\text{ such that } \norm{ \sum_{n=1}^\infty f_n - f}_\infty \to 0 \] In other words, the series converges uniformly. > Slogan: Convergence of the sup norms implies uniform convergence" ::: :::{.remark} The \(M\) in the name comes from defining $\sup\theset{f_k(x)} \da M_n$ and requiring $\sum \abs{M_n} < \infty$. ::: ### Checklist - Do the terms tend to zero? - $a_i \not\to 0 \implies \sum a_i = \infty$. - Can check with L'Hopital's rule - There are exactly 6 tests at our disposal: - Comparison, root, ratio, integral, limit, alternating - Is the series alternating? - If so, does $a_n \downarrow 0$? - If so, **convergent** - Is this series bounded above by a known convergent series? - \(p\) series with \(p>1\), i.e. : $\sum a_n \leq \sum \frac{1}{n^p} < \infty$ - Geometric series with $\abs{x} < 1$, i.e. $\sum a_n \leq \sum x^n$ - Is this series bounded below by a known divergent series? - \(p\) series with $p\leq 1$, i.e. $\infty = \sum \frac{1}{n^p} \leq \sum a_i$ - Are the ratios strictly less than or greater than 1? - $<1 \implies$ **convergent** - $>1 \implies$ **convergent** - Does the integral analogue converge? - Integral converges $\iff$ sum converges - Try the root test - $<1 \implies$ **convergent** - $>1 \implies$ **convergent** - Try the limit test - Attempt to divide each term to obtain a known convergent/divergent series Some Pattern Recognition: - $(\text{stuff})!$: Ratio test (only test that will work with factorials!!) - $(\text{stuff})^n$: Root test or ratio test - Replace \(a_n\) with an \(f(x)\) that's easy to integrate - integral test - \(p(x)\) or $\sqrt{p(x)}$: comparison or limit test ## Radius of Convergence :::{.proposition title="Finding the radius of convergence"} Use the fact that \[ \lim_{k\to\infty} \abs{\frac{a_{k+1}x^{k+1}}{a_kx^k}} = \abs{x}\lim_{k\to\infty} \abs{\frac{a_{k+1}}{a_k}} < 1 \implies \sum a_k x^k < \infty ,\] so take $L \da \lim_{k\to\infty} \frac{a_{k+1}}{a_k}$ and then obtain the radius as \[ R = \frac{1}{L} = \lim_{k\to\infty} {a_k \over a_{k+1}} \] ::: :::{.remark} \envlist - Note $L=0 \implies$ absolutely convergent everywhere - $L = \infty \implies$ convergent only at $x=0$. - Also need to check endpoints \(R, -R\) manually. :::