\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
deepdream of a sidewalk
Show Answer
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\groupAdd}[1] { +_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \require{physics} \require{ams} \require{mathtools} \)
Math and science::Analysis::Tao::07. Series

Series laws Ⅰ: finite series laws

1. Continuation. Let \( m \le n < p \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \le i \le p \). The we have

[\(\sum_{i=m}^{n} a_i + \sum_{i=n+1}^{p} a_i = ? \)]

2. Indexing shift. Let \( m \le n \) be integers, \( k \) be another integer, and let \( a_i \) be a real number assigned to each integer \( m \le i \le n \). Then we have

\[ \sum_{i=m}^{n}a_i = \sum_{j=m+k}^{n+k}[...] \]

This one above, I actually had trouble proving. It is almost too obvious.

3. [...]. Let \( m \le n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \le i \le n \). Then we have

\[ \sum_{i=m}^{n}(a_i + b_i) = \left( \sum_{i=m}^{n}a_i \right) + \left( \sum_{i=m}^{n}b_i \right) \]

4. [...]. Let \( m \le n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \le i \le n \), and let \( c \) be another real number. Then we have

\[ \sum_{i=m}^{n}(c a_i) = c \left( \sum_{i=m}^{n}a_i \right) \]

5. [...]. Let \( m \le n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \le i \le n \). Then we have

\[ \left| \sum_{i=m}^{n}a_i \right| \le \sum_{i=m}^{n}|a_i| \]

6. [...]. Let \( m \le n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \le i \le n \). Suppose that \( a_i \le b_i \) for all \( m \le i \le n \). Then we have

\[ \sum_{i=m}^{n}a_i \le \sum_{i=m}^{n}b_i \]

finite series → finite sets → infinite series → infinite sets (absolutely convergent series)