Sequences, series, summability Proposition

Limits preserve non-strict inequalities

Builds on Limit of a sequence Epsilon-N definition of the limit of a sequence

Statement

Let $(a_n)$ and $(b_n)$ be real sequences that converge to real numbers $a$ and $b$ . Assume that there is a natural number $N$ such that $a_n \le b_n$ for all $n \ge N$ . Then $a \le b$ .

Remarks

Strict inequalities do not pass to the limit. For example, $0 < 1/n$ for all $n \ge 1$ , but $\lim 1/n = 0$ . With a constant sequence $b_n = c$ , it says: if $a_n \le c$ for large $n$ , then $\lim a_n \le c$ . The proof is by contradiction. If $a > b$ , take $\varepsilon = (a - b)/2$ . Then for large $n$ we get $b_n < a_n$ .

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Sequence definition
Real sequence definition
Limit of a sequence definition
Epsilon-N definition of the limit of a sequence statement
Limits preserve non-strict inequalities statement

Tag	`0N32V0E3XJ`
URI	`m2://0N32V0E3XJ`
Kind	statement, proposition
Writing	W0 stub
MSC	40A05
Status	This copy of the site has no build data. Run `m2 status 0N32V0E3XJ`.
Uses	Limit of a sequence, Epsilon-N definition of the limit of a sequence

Limits preserve non-strict inequalities

Statement

Remarks

Reading path

Further reading

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