Question

Consider the sequence of sums $\frac{1}{3},\frac{1}{3}+\frac{1}{9},\frac{1}{3}+\frac{1}{9}+\frac{1}{27},\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81},\dots$

(a) What happens to the terms of this sequence of sums as k gets larger and larger?

(b) Find a sufficiently large value of k which will guarantee that every term past the kth term of this sequence of sums is in the interval (0.49999, 0.5).

Short answer

(a)$\underset{k\to \mathrm{\infty }}{lim} \left(\frac{1}{3^k}+\frac{1}{3^{k-1}}+\frac{1}{3^{k-2}}+\cdots +\frac{1}{3}\right)=\mathrm{\infty }$

(b)$0.4974$

Step-by-step solution

Part (a) Step 1. Given information

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

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 which will guarantee that every term past the k^th term of this sequence of sums is in the interval (0.49999, 0.5).

Part (a) Step 2. Terms of the sequence

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

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

$\underset{k\to \mathrm{\infty }}{lim} \left(\frac{1}{3^k}+\frac{1}{3^{k-1}}+\frac{1}{3^{k-2}}+\cdots +\frac{1}{3}\right)=\mathrm{\infty }$

Part (b) Step 1. Value of k

For every k>5, the terms will be :

$\begin{array}{r}\frac{1}{3^5}+\frac{1}{3^{5-1}}+\frac{1}{3^{5-2}}+\frac{1}{3^{5-3}}+\frac{1}{3}\\ =\frac{1}{3^5}+\frac{1}{3^4}+\frac{1}{3^3}+\frac{1}{3^2}+\frac{1}{3}\\ \approx 0.00412+0.0123+0.037+0.111+0.333\\ \approx 0.4974\end{array}$