How does monotonicity trap subsequent terms within the epsilon neighborhood?

Q: How does monotonicity trap subsequent terms within the epsilon neighborhood?

Monotonicity ensures that once a term $a_N$ exceeds $L - \varepsilon$, all subsequent terms $a_n$ (for $n > N$) are greater than or equal to $a_N$. Since the sequence is also bounded above by $L$, these terms satisfy $L - \varepsilon < a_n \le L$. This confines all tail terms to the interval $(L - \varepsilon, L]$, which is the epsilon neighborhood of the supremum $L$.

Monotonicity ensures that once a term

a_N

exceeds

L - \varepsilon

, all subsequent terms

a_n

(for

n > N

) are greater than or equal to

a_N

. Since the sequence is also bounded above by

L

, these terms satisfy

L - \varepsilon < a_n \le L

. This confines all tail terms to the interval

(L - \varepsilon, L]

, which is the epsilon neighborhood of the supremum

L

Conditions

The sequence is monotone increasing.
There exists an index $N$ such that $a_N > L - \varepsilon$ .
$L$ is an upper bound for the sequence.

Reasoning, step by step

Identify the term $a_N$ that exceeds $L - \varepsilon$ (guaranteed by the supremum property).
Apply the monotonicity condition: for all $n > N$ , $a_n \ge a_N$ .
Combine this with the lower bound: $a_n \ge a_N > L - \varepsilon$ .
Apply the global upper bound: $a_n \le L$ for all $n$ .
Conclude that for all $n > N$ , $L - \varepsilon < a_n \le L$ .

Example

The video states: 'Utilizing the monotonicity condition, if $a_N > L - ε$ , then every subsequent term $a_n$ (for $n > N$ ) must also satisfy $a_n \ge a_N > L - ε$ .' This creates a 'trap' where terms cannot fall back below $L - \varepsilon$ .

Common misconceptions

Thinking that monotonicity is not needed if the sequence is bounded.
Believing that terms can oscillate around $L - \varepsilon$ even in a monotone sequence.
Confusing the direction of the inequality in the monotonicity step.