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Sequences and Series /
Sequences and Series.tex
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\documentclass[12pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} \begin{align*} \text{Sequences:}\quad&\text{Ordered numbers following a rules. The terms are denoted by}\quad T_1, T_2, \ldots, T_n, \ldots\\ &\text{(Common notations: }a=T_1\quad\text{and}\quad l=T_n\text{)}\\ \text{Series:}\quad&\text{Sum of terms in a Sequence:}\quad S_n=\sum_{k=1}^{n}T_k=T_1+T_2+\ldots+T_{n-1}+T_n\\ \text{Arithmetic }&\text{Progressions (APs):}\\ &T_n-T_{n-1}=d\:,\quad\text{which is called the \it Common Difference.}\\ &\text{A simple test:}\quad\text{The sequence is an AP if}\quad T_3-T_2=T_2-T_1=d\:,\quad\text{or simply}\quad T_2=\frac{T_3+T_1}{2}\:.\\ \\ \because\quad&T_n-T_{n-k}=kd\\ \quad&T_n-T_1=T_n-T_{n-(n-1)}=(n-1)\:d\\ &T_n-a=(n-1)\:d\\ \therefore\quad&\boxed{T_n=a+(n-1)\:d}\quad\text{where $T_n$ is the ``n''th term.}\\ \text{or}\quad&l=a+(n-1)\:d\\ \\ &T_{k+2}-T_{k+1}=T_{k+1}-T_k\:(=d)\text{ for any positive integer }k\leq n-2\quad\Leftrightarrow\quad T\text{ is in }AP.\\ &\frac{T_k+T_{k+2}}{2}=T_{k+1},\text{ which is called the arithmetic mean of }T_k\text{ and }T_{k+2}.\\ \text{i.e.}\quad&\boxed{\text{The arithmetic mean of $a$ and $b$ is }\frac{a+b}{2}\:.}\\ \\ S_n&=\sum_{k=1}^{n}T_k=T_1+T_2+\ldots+T_{n-1}+T_n\\ &=a+(a+d)+(a+2d)+\ldots+[a+(n-2)d]+[a+(n-1)d]\\ &=na+d\:[1+2+\ldots+(n-2)+(n-1)]\\ &=na+d\:\frac{n(n-1)}{2}\\ &=\tfrac{\:1\:}{2}n\big[2a+(n-1)d\big]\\ \therefore\:&\boxed{S_n=\tfrac{\:1\:}{2}n\big[2a+(n-1)d\big]}\quad\text{where $S_n$ is the sum of the first ``n''th terms of the sequence.}\\ \\ l&=a+(n-1)\:d\\ a+l&=2a+(n-1)\:d\\ \therefore\:&\boxed{S_n=\tfrac{\:1\:}{2}\:n\:(a+l)}\\ \end{align*} \begin{align*} \text{Geometric }&\text{Progressions (GPs):}\\ &\frac{T_n}{T_{n-1}}=r\:,\quad\text{which is called the \it Common Ratio.}\\ \\ \because\quad&\frac{T_n}{T_{n-k}}=r^k\\ \quad&\frac{T_n}{T_1}=\frac{T_n}{T_{n-(n-1)}}=r^{n-1}\\ &\frac{T_n}{a}=r^{n-1}\\ \therefore\quad&\boxed{T_n=ar^{n-1}}\\ \text{or}\quad&l=ar^{n-1}\\ \\ &\frac{T_{k+2}}{T_{k+1}}=\frac{T_{k+1}}{T_k}\:(=r)\text{ for any positive integer }k\leq n-2\quad\Leftrightarrow\quad T\text{ is in }GP.\\ &\sqrt{T_k\cdot T_{k+2}}=\pm T_{k+1},\text{ which is called the geometric mean of }T_k\text{ and }T_{k+2}.\\ \text{i.e.}\quad&\boxed{\text{The geometric mean of $a$ and $b$ is}\quad\sqrt{ab}\quad\text{or}\quad-\sqrt{ab}\:.}\\ \\ S_n&=\sum_{k=1}^{n}T_k=T_1+T_2+\ldots+T_{n-1}+T_n\\ &=a+ar+ar^2+\ldots+ar^{n-2}+ar^{n-1}\\ &=a\:(1+r+r^2+\ldots+r^{n-2}+r^{n-1})\\ &=\frac{a(1-r^n)}{1-r}\\ \therefore\:&\boxed{S_n=\frac{a(1-r^n)}{1-r}}\quad\Big(\text{or }\frac{a(r^n-1)}{r-1}\text{ when }r>1\:\ldots\text{ if you really hate negatives.}\Big)\\ \text{and}\quad&\boxed{S_\infty=\frac{a}{1-r}\quad\text{where }|r|<1}\\ \\ l&=ar^{n-1}\\ a-lr&=a-ar^n=a\:1-r^n)\\ \therefore\:&\boxed{S_n=\frac{a-rl}{1-r}}\\ \\ \text{If $T_k$ is }&\text{in AP with common difference $d$, then $a^{T_k}$ is in GP with common ratio $a^d$.}\\ &\quad\left(a^{T_{k+1}}=a^{T_k +d}=a^{T_k}\cdot a^d\right)\\ \\ \text{If $T_k$ is }&\text{in GP with common ratio $r$, then $\log_a T_k$ is in AP with common difference $\log_a r$.}\\ &\quad\left(\log_a T_{k+1}=\log_a(T_k\cdot r)=\log_a T_k +\log_a r\right)\\ \end{align*} \end{document}