How do the properties of the supremum and monotonicity combine to trap sequence terms within the interval (L - ε, L]?

Conditions

• Sequence $\{a_n\}$ is monotone increasing.
• Sequence $\{a_n\}$ is bounded above by $L$.
• $\epsilon > 0$ is given.

Reasoning, step by step

1. Use the definition of supremum to find $N$ such that $a_N > L - \epsilon$.
2. Use monotonicity to establish $a_n \ge a_N$ for all $n > N$.
3. Combine steps to get $a_n > L - \epsilon$ for all $n > N$.
4. Use the upper bound property to state $a_n \le L$ for all $n$.
5. Merge inequalities to conclude $L - \epsilon < a_n \le L$ for $n > N$.

Example

The script derives: 'Combined with the global upper bound $a_n \le L$, we derive the inequality $L - ε < a_n \le L$. This confirms that beyond index N, all terms lie within the epsilon neighborhood of L.'

Common misconceptions

• Forgetting that the lower bound condition only applies for $n > N$, not necessarily for all $n$.
• Assuming $a_n$ equals $L$ eventually; the inequality allows $a_n$ to be strictly less than $L$.

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Explore next

• How does this prove the limit is exactly L using the epsilon-N definition?
• What changes if the sequence is not monotone?