Q. 9 - Calculus

Question

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. 

Step-by-Step Solution

Answer

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

(b) $0.0001524$

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:

$\underset{k\to \mathrm{\infty }}{lim} \frac{1}{3^k}=0$

Part (b) Step 1. Value of k

For every k>8, the terms will be :

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