// Numbas version: exam_results_page_options {"name": "Monotonic sequence theorem", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"description": "

Test whether a student knows the monotonic sequence theorem.

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The monotonic sequence theorem is \"Every bounded, monotonic sequence is convergent\". Which can be broken up into two cases:

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• If $\\{a_n\\}$ is monotonically non-decreasing (i.e. $a_{n} \\le a_{n+1}$ for all $n$) and bounded above (i.e. there exists a number $M$ such that $a_n\\le M$ for all $n$) then the sequence converges (i.e. there exists an $L$ such that as $n\\rightarrow \\infty$, $a_n\\rightarrow L$)

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• If $\\{a_n\\}$ is monotonically non-increasing (i.e. $a_{n} \\ge a_{n+1}$ for all $n$) and bounded below (i.e. there exists a number $m$ such that $a_n\\ge m$ for all $n$) then the sequence converges (i.e. there exists an $L$ such that as $n\\rightarrow \\infty$, $a_n\\rightarrow L$)

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Also, a sequence is called 'bounded' if it is both bounded above and bounded below.

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Notice, if $\\{a_n\\}$ is monotonically non-increasing and bounded above, or if $\\{a_n\\}$ is monotonically non-decreasing and bounded below, the theorem isn't useful since the sequence could still diverge to $-\\infty$ or $\\infty$ (respectively) or converge.

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a)

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By the monotonic sequence theorem, the sequence converges.

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The monotonic sequence theorem is not applicable here.

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b)

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The sequence $a_n=${bdict[\"a_n\"]} for $n\\ge1$ is monotonically non-decreasing (notice its derivative is positive for all $n\\ge 1$) and is bounded (i.e. bounded above and below).   The sequence $a_n=${bdict[\"a_n\"]} for $n\\ge1$ is monotonically non-increasing (notice its derivative is negative for all $n\\ge 1$) and is bounded (i.e. bounded above and below).   The sequence $a_n=${bdict[\"a_n\"]} for $n\\ge1$ is monotonically non-decreasing (notice its derivative is positive for all $n\\ge 1$) but is not bounded.   The sequence $a_n=${bdict[\"a_n\"]} for $n\\ge1$ is monotonically non-increasing (notice its derivative is negative for all $n\\ge 1$) but is not bounded.   The sequence $a_n=${bdict[\"a_n\"]} for $n\\ge1$ is not monotonic (notice it oscillates) and is not bounded.   The sequence $a_n=${bdict[\"a_n\"]} for $n\\ge1$ is not monotonic (notice it oscillates) but is bounded (i.e. bounded above and below).

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Therefore, by the monotonic sequence theorem, this sequence converges. Therefore the monotonic sequence theorem is not applicable to this sequence.

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This question is about the monotonic sequence theorem.

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The sequence converges.

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The sequence diverges.

", "

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Suppose $\\{a_n\\}$ is a non-decreasing sequence which is bounded above.

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Suppose $\\{a_n\\}$ is a non-increasing sequence which is bounded below.

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Suppose $\\{a_n\\}$ is a non-decreasing sequence which is bounded.

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Suppose $\\{a_n\\}$ is a non-increasing sequence which is bounded.

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Suppose $\\{a_n\\}$ is a non-increasing sequence which is bounded above.

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Suppose $\\{a_n\\}$ is a non-decreasing sequence which is bounded below.

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Suppose $\\{a_n\\}$ is an oscillating sequence which is bounded.

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What does the monotonic sequence theorem tell us about the sequence $\\{a_n\\}$?

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implies that this sequence is convergent

", "

does not apply to this sequence

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monotonically non-decreasing

", "

monotonically non-increasing

", "

not monotonic

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bounded

", "

not bounded

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Consider the sequence $a_n=${bdict[\"a_n\"]} for $n\\ge1$.

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The monotonic sequence theorem [[0]] since it is [[1]] and [[2]].

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do more combinations, maybe list both bounds?

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