Chapter B, Problem 29AYU

In Problems 29-32, find the first five terms in the sequence of partial sums for each series.

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{\left(\frac{1}{3}\right)}^{k-1}}$

To determine

To find: The first five terms in the sequence of partial sums for each series.

Answer to Problem 29AYU

Solution:

The first five terms in the sequence of partial sums for each series is $1, \left(\frac{4}{3}\right), \left(\frac{13}{9}\right), \left(\frac{40}{27}\right), \left(\frac{121}{81}\right)$.

Explanation of Solution

Given:

${\displaystyle {\sum }_{k = 1}^{\infty }{\left(\frac{1}{3}\right)}^{k - 1}}$

Calculation:

$\begin{array}{l}{\displaystyle {\sum }_{k = 1}^{\infty }{\left(\frac{1}{3}\right)}^{k - 1} = {\left(\frac{1}{3}\right)}^{1 - 1} + {\left(\frac{1}{3}\right)}^{2 - 1} + {\left(\frac{1}{3}\right)}^{3 - 1} + {\left(\frac{1}{3}\right)}^{3 - 1} + {\left(\frac{1}{3}\right)}^{5 - 1} + {\left(\frac{1}{3}\right)}^{6 - 1} + {\left(\frac{1}{3}\right)}^{7 - 1}\dots \dots \dots \dots \dots \dots }\\ = 1 + \left(\frac{1}{3}\right) + {\left(\frac{1}{3}\right)}^2 + {\left(\frac{1}{3}\right)}^3 + {\left(\frac{1}{3}\right)}^4 + {\left(\frac{1}{3}\right)}^5 + {\left(\frac{1}{3}\right)}^6 + \cdots \dots \dots \dots \dots \dots \\ = 1 + \left(\frac{1}{3}\right) + \left(\frac{1}{9}\right) + \left(\frac{1}{27}\right) + \left(\frac{1}{81}\right) + {\left(\frac{1}{3}\right)}^5 + {\left(\frac{1}{3}\right)}^6 + \cdots \dots \dots \dots \dots \dots \\ = 1 + \left(\frac{1}{3}\right) + \left(\frac{1}{9}\right) + \left(\frac{1}{27}\right) + \left(\frac{1}{81}\right) + \cdots \dots \mathrm{..}\end{array}$

The first five terms in the sequence of partial sums for each series is

$\begin{array}{l}1\\ 1 + \left(\frac{1}{3}\right) = \left(\frac{4}{3}\right)\\ 1 + \left(\frac{1}{3}\right) + \left(\frac{1}{9}\right) = \left(\frac{13}{9}\right)\\ 1 + \left(\frac{1}{3}\right) + \left(\frac{1}{9}\right) + \left(\frac{1}{27}\right) = \left(\frac{40}{27}\right)\\ 1 + \left(\frac{1}{3}\right) + \left(\frac{1}{9}\right) + \left(\frac{1}{27}\right) + \left(\frac{1}{81}\right) = \left(\frac{121}{81}\right)\end{array}$

The first five terms in the sequence of partial sums for each series is $1, \left(\frac{4}{3}\right), \left(\frac{13}{9}\right), \left(\frac{40}{27}\right), \left(\frac{121}{81}\right)$.