1：数列1

\(设 \left\{a_n\right\} 有界，记：\)

\[{\overline{a}}_n=\sup\left\{a_n,a_{n+1},\cdots\right\}\quad{\underline{a}}_n=\inf\left\{a_n,a_{n+1},\cdots\right\} \]

\(则\left\{{\overline{a}}_n\right\} 和\left\{{\underline{a}}_n\right\} 极限存在，且数列\left\{a_n\right\} 收敛的充要条件是 \left\{{\overline{a}}_n\right\} 和\left\{{\underline{a}}_n\right\} 的极限相等。\)

\(证明：\)

\[\because{\overline{a}}_n=\sup\left\{a_i\mid i\geq{n}\right\}\geq a_i \quad \forall i\geq{n+1} \]

\[\therefore{\overline{a}}_{n}为\left\{a_i\mid i\geq{n+1}\right\}的上界 \]

\[\therefore {\overline{a}}_n\geq\sup \left\{a_i\mid i\geq{n+1}\right\}={\overline{a}}_{n+1} \]

\[\therefore\left\{{\overline{a}}_n\right\} \downarrow\quad 同理\left\{{\underline{a}}_n\right\} \uparrow \]

\[\therefore\left\{{\overline{a}}_n\right\}与\left\{{\underline{a}}_n\right\}均单调有界，从而均存在极限。 \]

\(充分性：\)

\[\because{\underline{a}}_n\leq a_n \leq {\overline{a}}_n，结合夹逼原理可得 \left\{a_n\right\} 收敛。 \]

\(必要性：\)

\[记\lim a_n=a，根据定义有: \]

\[\left(\forall\varepsilon>0\right)\left(\exists N\in \textbf{N}\right)\left(n>N\right)\left(a-\frac{\varepsilon}{2}<a_n<a+\frac{\varepsilon}{2}\right) \]

\[故对于集合\left\{ a_n\mid n>N \right\}而言，a+\frac{\varepsilon}{2}是它的一个上界。 \]

\[{\therefore\overline{a}}_{N+1}=\sup\left\{a_n\mid n>N\right\}\leq a+\frac{\varepsilon}{2}< a+\varepsilon \]

\[\because\left\{ {\overline{a}}_n \right\}\downarrow \qquad \therefore n>N时\qquad{\overline{a}}_n<a+\varepsilon \]

\[同理 n>N时 \qquad {\underline{a}}_n>a-\varepsilon \]

\[\therefore\left(\forall\varepsilon>0\right)\left(\exists N\in \textbf{N}\right)\left(n>N\right)\left(a-\varepsilon<{\underline{a}}_n\leq {\overline{a}}_n<a+\varepsilon\right) \]

\[从而根据极限的定义我们有：\lim {\underline{a}}_n=a=\lim {\overline{a}}_n。\qquad\blacksquare \]