When is the inequality L - ε < a_n ≤ L valid for the sequence terms?

Q: When is the inequality L - ε < a_n ≤ L valid for the sequence terms?

This double inequality is valid for all indices $n$ such that $n > N$, where $N$ is the specific index guaranteed by the supremum property such that $a_N > L - \epsilon$. Before $N$, terms may be less than or equal to $L - \epsilon$. After $N$, monotonicity forces them to stay above $L - \epsilon$, while the upper bound keeps them at or below $L$.

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When is the inequality $L - ε < a_n \le L$ valid for the sequence terms?

This double inequality is valid for all indices

n

such that

n > N

, where

N

is the specific index guaranteed by the supremum property such that

a_N > L - \epsilon

. Before

N

, terms may be less than or equal to

L - \epsilon

. After

N

, monotonicity forces them to stay above

L - \epsilon

, while the upper bound keeps them at or below

L

Conditions

$N$ is defined such that $a_N > L - \epsilon$ .
$n$ is an integer index.
Sequence is monotone increasing and bounded above by $L$ .

Reasoning, step by step

Identify the threshold index $N$ derived from the supremum definition.
Check the condition $n > N$ .
Verify that for $n > N$ , $a_n \ge a_N > L - \epsilon$ due to monotonicity.
Verify that $a_n \le L$ holds for all $n$ due to the upper bound.
Conclude validity specifically for the tail of the sequence ( $n > N$ ).

Example

The script states: 'This confirms that beyond index N, all terms lie within the epsilon neighborhood of L... We derive the inequality $L - ε < a_n \le L$ .' Implicitly, this derivation relies on $n > N$ .

Common misconceptions

Assuming the inequality holds for $n=1$ or small $n$ .
Thinking the inequality implies $a_n$ is constant; it just bounds the range.

Watch the explanation

BilibiliMonotone convergence from the supremum principle

0:24 – 0:50Watch this moment ↗

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