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Let akbe a sequence. Prove Theorem 7.6 (a) along with the following variations:

(a) Show that when role="math" localid="1649277359535" ak-1-ak≥ 0 for every k ≥ 1, the sequence is increasing.

(b) Show that when ak-1-ak> 0 for every k ≥ 1, the sequence is strictly increasing.

(c) Show that when role="math" localid="1649277346412" ak-1-ak≤ 0 for every k ≥ 1, the sequence is decreasing.

(d) Show that when ak-1-ak< 0 for every k ≥ 1, the sequence is strictly decreasing.

Short Answer

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Proved

Step by step solution

01

Step 1. Given

Consider the sequenceak

02

Part (a) Step 2. Explanation

Since,ak+1-ak≥0,nowallthetermsofthesequencearepositive,thereforeakisalsopositive.Now,addingtheinequalityak+1-ak≥0withak.Thesignoftheinequalitywillnotchangeasakispositive.Now,ak+1-ak+ak≥0+akak+1≥akfork≥1Hence,bedefinitionthesequenceisincreasingsequence.

03

Part(b) Step 3. Explanation

Since,ak+1-ak>0,nowallthetermsofthesequencearepositive,thereforeakisalsopositive.Now,addingtheinequalityak+1-ak>0withak.Thesignoftheinequalitywillnotchangeasakispositive.Now,ak+1-ak+ak>0+akak+1>akfork≥1Hence,bedefinitionthesequenceisstrictlyincreasingsequence.

04

Part(c) Step 4. Explanation

Since,ak+1-ak≤0,nowallthetermsofthesequencearepositive,thereforeakisalsopositive.Now,addingtheinequalityak+1-ak≤0withak.Thesignoftheinequalitywillnotchangeasakispositive.Now,ak+1-ak+ak≤0+akak+1≤akfork≥1Hence,bedefinitionthesequenceisdecreasingsequence.

05

Part(d) Step 5. Explanation

Since,ak+1-ak<0,nowallthetermsofthesequencearepositive,thereforeakisalsopositive.Now,addingtheinequalityak+1-ak<0withak.Thesignoftheinequalitywillnotchangeasakispositive.Now,ak+1-ak+ak<0+akak+1<akfork≥1Hence,bedefinitionthesequenceisstrictlydecreasingsequence.

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