Chapter 1: Q. 9 (page 86)

Consider the sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}, \dots, \frac{1}{3^{k}}, \dots$
(a) What happens to the terms of this sequence as k gets larger and larger? Express your answer in limit notation.
(b) Find a sufficiently large value of k so that every term past the kth term of this sequence will be less than 0.0001.

Short Answer

Expert verified

(a) $lim_{k \to \infty} \frac{1}{3^{k}} = 0$

(b) $0.0001524$

Step by step solution

Part (a) Step 1. Given information

Given is the sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}, \dots, \frac{1}{3^{k}}, \dots$

We have to explain What happens to the terms of this sequence as k gets larger and larger and Find a sufficiently large value of k so that every term past the kth term of this sequence will be less than 0.0001.

Part (a) Step 2. Terms of the sequence

From the sequence, we see that, as k gets larger and large, the terms get smaller and smaller.

Therefore, in limit expression it can be written as below:

$lim_{k \to \infty} \frac{1}{3^{k}} = 0$

Part (b) Step 1. Value of k

For every k>8, the terms will be :

$\begin{aligned} \frac{1}{3^{8}} = \frac{1}{6561} \\ \approx 0.0001524 \end{aligned}$

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Short Answer

Step by step solution

Part (a) Step 1. Given information

Part (a) Step 2. Terms of the sequence

Part (b) Step 1. Value of k

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Consider the sequence13,19,127,181,…,13k,…(a) What happens to the terms of this sequence as k gets larger and larger? Express your answer in limit notation.(b) Find a sufficiently large value of k so that every term past the kth term of this sequence will be less than 0.0001.

Short Answer

Step by step solution

Part (a) Step 1. Given information

Part (a) Step 2. Terms of the sequence

Part (b) Step 1. Value of k

One App. One Place for Learning.

Most popular questions from this chapter

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Discrete Mathematics

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Study anywhere. Anytime. Across all devices.