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When applying the Monotone Convergence Principle to the right endpoints {bn}\{b_n\}, what specific properties must the sequence satisfy?

The sequence {bn}\{b_n\} must be monotone decreasing (non-increasing) and bounded below. In the context of nested intervals, bn+1≤bnb_{n+1} \le b_n follows from the containment [an+1,bn+1]⊆[an,bn][a_{n+1}, b_{n+1}] \subseteq [a_n, b_n], ensuring monotonicity. Boundedness below is ensured because bn≥a1b_n \ge a_1 for all nn (since a1a_1 is the smallest possible left bound, and actually bn≥an≥a1b_n \ge a_n \ge a_1 is not quite right, rather bn≥anb_n \ge a_n and ana_n is increasing, so bnb_n is bounded below by any aka_k, specifically a1a_1 works as a loose lower bound, or more precisely, the entire sequence is trapped between a1a_1 and b1b_1). Actually, simpler: bn≥an≥a1b_n \ge a_n \ge a_1, so bnb_n is bounded below by a1a_1.

Conditions

  • Intervals are nested: [an+1,bn+1]⊆[an,bn][a_{n+1}, b_{n+1}] \subseteq [a_n, b_n].

Reasoning, step by step

  1. Verify monotonicity: From inclusion, bn+1≤bnb_{n+1} \le b_n.
  2. Verify boundedness: Find a constant MM such that bn≥Mb_n \ge M for all nn.
  3. Observe that bn≥anb_n \ge a_n and an≥a1a_n \ge a_1 (since {an}\{a_n\} increases from a1a_1).
  4. Therefore, bn≥a1b_n \ge a_1 for all nn.
  5. Conclude that {bn}\{b_n\} is decreasing and bounded below, hence convergent.

Example

Script states: 'the sequence of right endpoints {bn}\{b_n\} is strictly decreasing and has a lower bound.'

Common misconceptions

  • Thinking {bn}\{b_n\} needs to be bounded above (it is naturally bounded by b1b_1).
  • Confusing the direction of monotonicity for right vs left endpoints.

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