\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \)
deepdream of
          a sidewalk
Show Answer
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \)
Math and science::Analysis::Tao::06. Limits of sequences

Monotone bounded sequences converge

Increasing, decreasing and monotone

A sequence \( (a_n)_{n=0}^{\infty} \) is increasing if \( a_n \le a_{n+1} \) for all \( n \in \mathbb{N} \) and decreasing if \( a_n \gt a_{n+1} \) for all \( n \in \mathbb{N} \). A sequence is monotone if it is either increasing or decreasing.

Monotone bounded sequences converge

If a sequence is [...] and [...], then it converges.

The easiest card ever. The proof is on the reverse; can you think of it? The proof is easy and obvious in retrospect, but it uses a property that I originally didn't consider using to solve the problem.