How does the monotonicity of an increasing sequence ensure that all terms beyond index N satisfy $a_n \ge a_N$?

Conditions

• The sequence $\{a_n\}$ is monotone increasing.
• $N$ is a fixed integer index.
• $n$ is any integer such that $n > N$.

Reasoning, step by step

1. Recall the definition of a monotone increasing sequence: $a_k \le a_{k+1}$ for all $k$.
2. Identify the specific index $N$ where $a_N > L - \epsilon$.
3. Apply the monotonicity property repeatedly for indices between $N$ and $n$.
4. Conclude that $a_n \ge a_N$ for any $n > N$.
5. Combine this with the lower bound to show $a_n > L - \epsilon$.

Example

If $a_5 = 0.9$ and the sequence is increasing, then $a_6 \ge 0.9$, $a_7 \ge 0.9$, etc. The script states: 'if $a_N > L - ε$, then every subsequent term $a_n$ (for $n > N$) must also satisfy $a_n \ge a_N > L - ε$.'

Common misconceptions

• Believing that later terms could dip below earlier terms in an increasing sequence.
• Confusing monotone increasing ($a_{n+1} \ge a_n$) with strictly increasing ($a_{n+1} > a_n$); the proof holds for both, but weak inequality is sufficient here.

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Explore next

• What happens if the sequence is decreasing instead?
• Why is the upper bound L necessary for the convergence proof?