Q. 94

Question

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

(a) Show that when 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 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. 

Step-by-Step Solution

Verified
Answer

Proved

1Step 1. Given

Consider the sequence ak

2Part (a) Step 2. Explanation

Since, ak+1-ak0, now all the terms of the sequence are positive, therefore ak is also positive.Now, adding the inequality ak+1-ak0 with ak. The sign of the inequality will not change as ak is positive.Now,ak+1-ak+ak0+akak+1ak for k1Hence, be definition the sequence is  increasing sequence.

3Part(b) Step 3. Explanation

Since, ak+1-ak>0, now all the terms of the sequence are positive, therefore ak is also positive.Now, adding the inequality ak+1-ak>0 with ak. The sign of the inequality will not change as ak is positive.Now,ak+1-ak+ak>0+akak+1>ak for k1Hence, be definition the sequence is strictly  increasing sequence.

4Part(c) Step 4. Explanation

Since, ak+1-ak0, now all the terms of the sequence are positive, therefore ak is also positive.Now, adding the inequality ak+1-ak0 with ak. The sign of the inequality will not change as ak is positive.Now,ak+1-ak+ak0+akak+1ak for k1Hence, be definition the sequence is  decreasing sequence.

5Part(d) Step 5. Explanation

Since, ak+1-ak<0, now all the terms of the sequence are positive, therefore ak is also positive.Now, adding the inequality ak+1-ak<0 with ak. The sign of the inequality will not change as ak is positive.Now,ak+1-ak+ak<0+akak+1<ak for k1Hence, be definition the sequence is strictly decreasing sequence.