\documentclass[../../script.tex]{subfiles} % !TEX root = ../../script.tex \begin{document} \section{Convergence of Series} \begin{defi} Let $\rcseqdef{x}$. Then the series \[ \series{k} x_k \] is the sequence of partial sums $\seq{s}$: \[ s_n = \series[n]{k} x_k \] If the series converges, then $\series{k}$ denotes the limit. \end{defi} \begin{thm}\label{thm:seriesnull} Let $\rcseqdef{x}$. Then \[ \series{n} x_n \text{ converges} \implies \seq{x} \text{ null sequence} \] \end{thm} \begin{proof} Let $s_n = \series{n} x_n$. This is a Cauchy series. Let $\epsilon > 0$. Then \begin{equation} \exists N \in \natn ~\forall n \ge N: ~~|s_{n+1} - s_n| = |x_{n+1}| < \epsilon \end{equation} \end{proof} \begin{eg}[Geometric series] Let $x \in \realn$ (or $\cmpln$). Then \[ \series{k} x^k \] converges if $|x| < 1$. (Why?) \end{eg} \begin{eg}[Harmonic series] This is a good example of why the inverse of \Cref{thm:seriesnull} does not hold. Consider \[ x_n = \frac{1}{n} \] This is a null sequence, but $\series{k}\frac{1}{k}$ does not converge. (Why?) \end{eg} \begin{lem} Let $\rcseqdef{x}$. Then \[ \series{k} x_n \text{ converges} \iff \sum_{k=N}^{\infty} x_n \text{ converges for some } N \in \natn \] \end{lem} \begin{proof} \reader \end{proof} \begin{thm}[Alternating series test]\label{thm:alttest} Let $\seq{x} \subset [0, \infty)$ be a monotonic decreasing null sequence. Then \[ \series{k} (-1)^k x_k \] converges, and \[ \left| \series{k} (-1)^k x_k - \series[N]{k} (-1)^k x_k \right| \le x_{N+1} \] \end{thm} \begin{proof} Let $s_n = \series[n]{k} (-1)^k x_n$, and define the sub sequences $a_n = s_{2n}$, $b_n = s_{2n+1}$. Then \begin{equation} a_{n+1} = s_{2n} - \underbrace{(x_{2n+1} - x_{2n+2})}_{\ge 0} \le s_{2n} = a_n \end{equation} Hence, $\seq{a}$ is monotonic decreasing. By the same argument, $\seq{b}$ is monotonic decreasing. Let $m, n \in \natn$ such that $m \le n$. Then \begin{equation}\label{eq:act} b_m \le b_n = a_n - x_{2n+1} \le a_n \le a_m \end{equation} Therefore $\seq{a}$, $\seq{b}$ are bounded. By \Cref{thm:monotone}, these sequence converge \begin{align} \seq{a} &\convinf a & \seq{b} &\convinf b \end{align} Furthermore \begin{equation} b_n - a_n = -x_{2n+1} \convinf 0 \implies a = b \end{equation} From \cref{eq:act} we know that \begin{equation} b_m \le b = a \le a_m \end{equation} So therefore \begin{align} |s_{2n} - a| &= a_n - a \le a_n - b_n = x_{2n+1} \\ |s_{2n+1} - a| &= b - b_n \le a_{m+1} - b_n = x_{2n+2} \end{align} \end{proof} \begin{eg}[Alternating harmonic series] \[ \begin{split} s = \series{k} (-1)^{k+1} \rec{k} &= 1 - \rec{2} + \rec{3} - \rec{4} + \rec{5} - \cdots \\ &= \left(1 - \rec{2}\right) - \rec{4} + \left(\rec{3} - \rec{6}\right) - \rec{8} + \left(\rec{5} - \rec{10}\right) - \rec{12} + \cdots \\ &= \rec{2} - \rec{4} + \rec{6} - \rec{8} + \rec{10} - \rec{12} + \cdots \\ &= \rec{2}\left(1 - \rec{2} + \rec{3} - \rec{4} + \rec{5} - \rec{6} + \cdots\right) \\ &= \rec{2}s \end{split} \] But $s \in \left[\rec{2}, 1\right]$, this is an example on why rearranging infinite sums can lead to weird results. \end{eg} \begin{rem}\leavevmode \begin{enumerate}[(i)] \item The convergence behaviour does not change if we rearrange finitely many terms. \item Associativity holds without restrictions \[ \series{k} x_k = \series{k} (x_{2k} + x_{2k-1}) \] \item Let $I$ be a set, and define \begin{align*} I &\longrightarrow \realn \\ i &\longmapsto a_i \end{align*} Consider the sum \[ \sum_{i \in I} a_i \] If $I$ is finite, there are no problems. However if $I$ is infinite then the solution of that sum can depend on the order of summation! \end{enumerate} \end{rem} \begin{defi} Let $\rcseqdef{x}$. The series $\series{k} x_k$ is said to converge absolutely if $\series{k} |x_k|$ converges. \end{defi} \begin{rem} Let $\seq{x} \subset [0, \infty)$. Then the sequence \[ s_n = \series[n]{k} x_k \] is monotonic increasing. If $\seq{s}$ is bounded it converges, if it is unbounded it diverges properly. The notation for absolute convergence is \[ \series{k} |x_k| < \infty \] \end{rem} \begin{lem}\label{lem:absolutebounded} Let $\series{k} x_k$ be a series. Then the following are all equivalent \begin{enumerate}[(i)] \item \[ \series{k} x_k \text{ converges absolutely} \] \item \[ \set[I \subset \natn \text{ finite}]{\sum_{k \in I} |x_k|} \text{ is bounded} \] \item \[ \forall \epsilon > 0 ~\exists I \subset \natn \text{ finite} ~\forall J \subset \natn \text{ finite}: ~~\sum_{k \in J \setminus I} |x_k| < \epsilon \] \end{enumerate} \end{lem} \begin{proof} To prove the equivalence of all of these statements, we will show that (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (i). This is sufficient. First we prove (i) $\implies$ (ii). Let \begin{equation} \series{n} |x_n| = k \in [0, \infty) \end{equation} Let $I \subset \natn$ be a finite set, and let $N = \max I$. Then \begin{equation} \sum_{n \in I} |x_n| \le \series[N]{n} |x_n| \leexpl{Monotony of the partial sums} \series{n} |x_n| \end{equation} Now to prove (ii) $\implies$ (iii), set \begin{equation} K := \set[I \subset \natn \finite]{\sum_{k \in I} |x_k|} \end{equation} Let $\epsilon > 0$. Then by definition of $\sup$ \begin{equation} \exists I \subset \natn \finite: ~~\sum_{k \in I} |x_k| > k - \epsilon \end{equation} Let $J \subset \natn \finite$. Then \begin{equation} k - \epsilon < \sum_{k \in I} |x_k| \le \sum_{k \in I \cup J} |x_k| \le K \end{equation} Hence \begin{equation} \sum_{k \in J \setminus I} |x_k| = \sum_{k \in I \cup J} |x_k| - \sum_{k \in I} |x_k| \le \epsilon \end{equation} Finally we show that (iii) $\implies$ (i). Choose $I \subset \natn \finite$ such that \begin{equation} \forall J \subset \natn \finite: ~~\sum_{k \in J \setminus I} |x_k| < 1 \end{equation} Then $\forall J \subset \natn \finite$ \begin{equation} \sum_{k \in J} |x_k| \le \sum_{k \in J \setminus I} |x_k| + \sum_{k \in I} |x_k| \le \sum_{k \in I} |x_k| + 1 \end{equation} Therefore $\series[n]{k} |x_k|$ is bounded and monotonic increasing, and hence it is converging. So $\series{k} |x_k| < \infty$. \end{proof} \begin{thm}\label{259} Every absolutely convergent series converges and the limit does not depend on the order of summation. \end{thm} \begin{proof} Let $\series{k} x_k$ be absolutely convergent and let $\epsilon > 0$. Choose $I \subset \natn \finite$ such that \begin{equation} \forall J \subset \natn: ~~\sum_{k \in I} |x_k| < \epsilon \end{equation} Choose $N = \max I$. Define the series \begin{equation} s_n = \series[n]{k} x_k \end{equation} Then for $n \le m \le N$ \begin{equation} |s_n - s_m| \le \sum_{k=m+1}^n |x_k| \le \sum_{k \in \set{1, \cdots, n} \setminus I} |x_k| < \epsilon \end{equation} Hence $s_n$ is a Cauchy sequence, so it converges. Let $\phi: \natn \rightarrow \natn$ be a bijective mapping. According to \Cref{lem:absolutebounded} the series $\series{k} x_{\phi(n)}$ converges absolutely. Let $\epsilon > 0$. According to the same Lemma \begin{equation} \exists I \subset \natn \finite ~\forall J \subset \natn \finite: ~~\sum_{k \in J \setminus I} |x_k| < \frac{\epsilon}{2} \end{equation} Choose $N \in \natn$ such that \begin{equation} I \subset \set{1, \cdots, N} \cap \set{\phi(1), \phi(2), \cdots, \phi(n)} \end{equation} Then for $n \ge N$ \begin{equation} \begin{split} \left|\series{k} x_k - \series[n]{k} x_{\phi(k)}\right| &= \left| \sum_{k \in \set{1, \cdots, N} \setminus I} x_k - \sum_{k \in \set{\phi(1), \cdots, \phi(n)} \setminus I} x_k \right| \\ &\le \sum_{k \in \set{1, \cdots, N} \setminus I} |x_k| + \sum_{k \in \set{\phi(1), \cdots, \phi(n)} \setminus I} |x_k| < \epsilon \end{split} \end{equation} Therefore \begin{equation} \limn\left( \series[n]{k} x_k - \series[n]{k} x_{\phi(k)} \right) = 0 \end{equation} \end{proof} \begin{thm} Let $\series{k} x_k$ be a converging series. Then \[ \left| \series{k} x_k \right| \le \series{k} |x_k| \] \end{thm} \begin{proof} \reader \end{proof} \begin{thm}[Direct comparison test] Let $\series{k} x_k$ be a series. If a converging series $\series{k} y_k$ exists with $|x_k| \le y_k$ for all sufficiently large $k$, then $\series{k} x_k$ converges absolutely. If a series $\series{k} z_k$ diverges with $0 \le z_k \le x_k$ for all sufficiently large $k$, then $\series{k} x_k$ diverges. \end{thm} \begin{proof} \begin{equation} \series[n]{k} |x_k| \le \series[n]{k} y_k \implies \series[n]{k} x_k \text{ bounded} \implbl{\cref{lem:absolutebounded}} \series{k} |x_k| < \infty \end{equation} \begin{equation} \series[n]{k} z_k \le \series[n]{k} x_k \implies \series{k} x_k \text{ unbounded} \end{equation} \end{proof} \begin{cor}[Ratio test] Let $\seq{x}$ be a sequence. If $\exists q \in (0, 1)$ such that \[ \left| \frac{x_{n+1}}{x_n} \right| \le q \] for a.e. $n \in \natn$, then $\series{k} x_k$ converges absolutely. If \[ \left| \frac{x_{n+1}}{x_n} \right| \ge 1 \] then the series diverges. \end{cor} \begin{proof} Let $q \in (0, 1)$ and choose $N \in \natn$ such that \begin{equation} \forall n \ge N: ~~\left|\frac{x_{n+1}}{x_n}\right| \le q \end{equation} Then \begin{equation} |x_{N+1}| \le q|x_N|, ~|x_{N+2}| \le q|x_{N+1}| \le q^2|x_N|, ~\cdots \end{equation} This means that \begin{equation} \series{k} |x_k| \le \series[N]{k} |x_k| + \sum_{k=N+1}^\infty q^{k-N} \cdot |x_N| < \infty \end{equation} Hence, $\series{k} x_k$ converges absolutely. Now choose $N \in \natn$ such that \begin{equation} \forall n \ge N: ~~\left|\frac{x_{n+1}}{x_n}\right| > 1 \end{equation} However this means that \begin{equation} |x_{n+1}| \ge |x_{n}| ~~\forall n \ge N \end{equation} So $\seq{x}$ is monotonic increasing and therefore not a null sequence. Hence $\series{k} x_k$ diverges. \end{proof} \begin{cor}[Root test] Let $\seq{x}$ be a sequence. If $\exists q \in (0, 1)$ such that \[ \sqrt[n]{|x_n|} \le q \] for a.e. $n \in \natn$, then $\series{k} x_k$ converges absolutely. If \[ \sqrt[n]{|x_n|} \ge 1 \] for all $n \in \natn$ then $\series{k} x_k$ diverges. \end{cor} \begin{proof} \reader \end{proof} \begin{rem} The previous tests can be summed up by the formulas \begin{align*} \limn \left|\frac{x_{n+1}}{x_n}\right| &< 1 & \limn \sqrt[n]{|x_n|} &< 1 \\ \limn \left|\frac{x_{n+1}}{x_n}\right| &> 1 & \limn \sqrt[n]{|x_n|} &> 1 \end{align*} for convergence and divergence respectively. If any of these limits is equal to $1$ then the test is inconclusive. \end{rem} \begin{eg} Let $z \in \cmpln$. Then \[ \exp(z) := \sum_{k=0}^\infty \frac{z^k}{k!} \] converges. To prove this, apply the ratio test: \[ \frac{|z|^{k+1} k!}{(k+1)! |z|^k} = \frac{|z|}{k+1} \conv{} 0 \] The function $\exp: \cmpln \rightarrow \cmpln$ is called the exponential function. \end{eg} \begin{rem}[Binomial coefficient] The binomial coefficient is defined as \begin{align*} {n\choose 0} &:= 1 & {n\choose k+1} &= {n\choose k} \cdot \frac{n-k}{k+1} \end{align*} and represents the number of ways one can choose $k$ objects from a set of $n$ objects. Some rules are \begin{enumerate}[(i)] \item \[{n\choose k} = 0 ~~\text{ if } k > n\] \item \[k \le n: ~~{n\choose k} = \frac{n!}{k!(n-k)!}\] \item \[{n\choose k} + {n\choose k-1} = {n+1\choose k}\] \item \[\forall x, y \in \cmpln: ~~(x+y)^n = \series[n]{k} {n\choose k} x^ky^{n-k}\] \end{enumerate} \end{rem} \begin{thm} \[ \forall u, v \in \cmpln: ~~\exp(u+v) = \exp(u)\cdot\exp(v) \] \end{thm} \begin{proof} \begin{equation} \begin{split} \exp(u)\cdot\exp(v) = \left(\sum_{n=0}^{\infty} \frac{u^n}{n!} \right) \cdot \left(\sum_{m=0}^{\infty} \frac{v^m}{m!} \right) &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{u^nv^m}{n!m!} \\ &= \sum_{l=0}^{\infty} \sum_{k=0}^{l} \frac{u^kv^{l-k}}{k!(l-k)!} \\ &=\sum_{l=0}^{\infty} \frac{(u+v)^l}{l!} \\ &= \exp(u+v) \end{split} \end{equation} \end{proof} \begin{rem} We define Euler's number as \[ e := \exp(1) \] We will also take note of the following rules $\forall x \in \cmpln, n \in \natn$ \[ \exp(0) = \exp(x)\exp(-x) = 1 \implies \exp(-x) = \frac{1}{\exp(x)} \] \[ \exp(nx) = \exp(x + x + x + \cdots + x) = \exp(x)^n \] \[ \exp(x)^{\frac{1}{n}} = \exp(\frac{x}{n}) \] Alternatively we can write \[ \exp(z) = e^z \] \end{rem} \begin{thm} Let $x, y \in \realn$. \begin{enumerate}[(i)] \item \[ x < y \implies \exp(x) < \exp(y) \] \item \[ \exp(x) > 0 ~~\forall x \in \realn \] \item \[ \exp(x) \ge 1 + x ~~\forall x \in \realn \] \item \[ \limn \frac{n^d}{\exp(n)} = 0 ~~\forall d \in \natn \] \end{enumerate} \end{thm} \begin{proof}\leavevmode \begin{enumerate}[(i)] \item \reader \item For $x \ge 0$ this is trivial. For $x < 0$ \begin{equation} \exp(x) = \frac{1}{\exp(-x)} > 0 \end{equation} \item For $x \ge 0$ this is trivial. For $x < 0$ \begin{equation} \sum_{k=0}^{\infty} \frac{x^k}{k!} \end{equation} is an alternating series, and therefore the statement follows from \Cref{thm:alttest}. \item Let $d \in \natn$. Then $\forall n \in \natn$ \begin{equation} 0 < \frac{n^d}{\exp(n)} < \frac{n^d}{\sum_{k=0}^{d+1} \frac{n^k}{k!}} \convinf 0 \end{equation} \end{enumerate} \end{proof} \begin{defi} Define \[ \sin, \cos: \realn \longrightarrow \realn \] as \begin{align*} \sin(x) &:= \Im(\exp(ix)) \\ \cos(x) &:= \Re(\exp(ix)) \end{align*} \end{defi} \begin{rem}\leavevmode \begin{enumerate}[(i)] \item Euler's formula \[ \exp(ix) = \cos(x) + i\sin(x) \] \item $\forall z \in \cmpln: ~~\overline{\exp(z)} = \exp(\bar{z})$ \[ |\exp(ix)|^2 = \exp(ix) \cdot \overline{\exp(ix)} = \exp(ix) \cdot \exp(-ix) = 1 \] Also: \[ 1 = \cos^2(x) + \sin^2(x) \] On the symmetry of $\cos$ and $\sin$: \[ \cos(-x) + i\sin(-x) = \exp(-ix) = \overline{\exp(ix)} = \cos(x) - i\sin(x) \] \item From \[ \exp(ix) = \sum_{k=0}^{\infty} \frac{(ix)^k}{k!} ~~~(i^0 = 1, i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, \cdots) \] follow the following series \begin{align*} \sin(x) &= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!} & \cos(x) &= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!} \end{align*} \item For $x \in \realn$ \[ \begin{split} \exp(i2x) &= \cos(2x) + i\sin(2x) \\ &= (\cos(x) + i\sin(x))^2 \\ &= \cos^2(x) - \sin^2(x) + 2i\sin(x)\cos(x) \end{split} \] By comparing the real and imaginary parts we get the following identities \begin{align*} \cos(2x) &= \cos^2(x) - \sin^2(x) \\ \sin(2x) &= 2\sin(x)\cos(x) \end{align*} \item Later we will show that $\cos$ as exactly one root in the interval $[0, 2]$. We define $\pi$ as the number in the interval $[0, 4]$ such that $\cos(\frac{\pi}{2}) = 0$. \[ \implies \sin(\frac{\pi}{2}) = \pm 1 \] $\cos$ and $\sin$ are $2\pi$-periodic. \end{enumerate} \end{rem} \begin{thm} $\forall z \in \cmpln$ \[ \limn \left(1 + \frac{z}{n}\right)^n = \limn \left(1 - \frac{z}{n}\right)^{-n} = \exp(z) \] \end{thm} \begin{proof} Without proof. \end{proof} \end{document}