Let akbe a sequence of positive terms. Prove Theorem 7.6 (b) along with the following variations:(a) Show that when ak+1ak≥1≥ 1 for every k ≥ 1, the sequence is increasing.(b) Show that when localid="1649277252156" ak+1ak>1 for every k ≥ 1, the sequence is strictly increasing.(c) Show that when localid="1649277261045" ak+1ak≤1for every k ≥ 1, the sequence is decreasing.(d) Show that when localid="1649277269413" ak+1ak<1 for every k ≥ 1, the...

Q. 95 - Calculus | StudyQuestionHub

Q. 95

Question

Let $\{a_{k}\}$ be a sequence of positive terms. Prove Theorem 7.6 (b) along with the following variations:

(a) Show that when $\frac{a_{k + 1}}{a_{k}} \geq 1$ ≥ 1 for every k ≥ 1, the sequence is increasing.

(b) Show that when $\frac{a_{k + 1}}{a_{k}} > 1$ for every k ≥ 1, the sequence is strictly increasing.

(d) Show that when $\frac{a_{k + 1}}{a_{k}} < 1$ for every k ≥ 1, the sequence is strictly decreasing.

Step-by-Step Solution

Verified

Answer

Proved.

1Step 1. Given

Let $\{a_{k}\}$ be a sequence of positive terms.

2Part (a) Step 2. Explanation

$Since, \frac{a_{k + 1}}{a_{k}} \geq 1, now all the terms of the sequence are positive, therefore a_{k} is also positive . Now, multiply the inequality \frac{a_{k + 1}}{a_{k}} \geq 1 with a_{k} . The sign of the inequality will not change as a_{k} is positive . Now, \frac{a_{k + 1}}{a_{k}} (a_{k}) \geq (a_{k}) a_{k + 1} \geq a_{k} for k \geq 1 Hence, be definition the sequence is increasing sequence .$

3Part (b) Step 3. Explanation

$Since, \frac{a_{k + 1}}{a_{k}} > 1, now all the terms of the sequence are positive, therefore a_{k} is also positive . Now, multiply the inequality \frac{a_{k + 1}}{a_{k}} > 1 with a_{k} . The sign of the inequality will not change as a_{k} is positive . Now, \frac{a_{k + 1}}{a_{k}} (a_{k}) > (a_{k}) a_{k + 1} > a_{k} for k \geq 1 Hence, be definition the sequence is Strictly increasing sequence .$

4Part(c) Step 4. Explanation

$Since, \frac{a_{k + 1}}{a_{k}} \leq 1, now all the terms of the sequence are positive, therefore a_{k} is also positive . Now, multiply the inequality \frac{a_{k + 1}}{a_{k}} \leq 1 with a_{k} . The sign of the inequality will not change as a_{k} is positive . Now, \frac{a_{k + 1}}{a_{k}} (a_{k}) \leq (a_{k}) a_{k + 1} \leq a_{k} for k \geq 1 Hence, be definition the sequence is d e creasing sequence .$

5Part (d) Step 5. Explanation

$Since, \frac{a_{k + 1}}{a_{k}} < 1, now all the terms of the sequence are positive, therefore a_{k} is also positive . Now, multiply the inequality \frac{a_{k + 1}}{a_{k}} < 1 with a_{k} . The sign of the inequality will not change as a_{k} is positive . Now, \frac{a_{k + 1}}{a_{k}} (a_{k}) < (a_{k}) a_{k + 1} < a_{k} for k \geq 1 Hence, be definition the sequence is strictly decreasing sequence .$

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