# Science:Infinite Series Module/Units/Unit 1/1.4 Properties of Convergent Series/1.4.02 Properties of Convergent Series Example

## Problem

Find the sum of the series

{\begin{aligned}\sum _{k=1}^{\infty }{\Big (}{\frac {4}{5^{k-1}}}-{\frac {2^{k}}{3^{k}}}{\Big )}.\end{aligned}} ## Complete Solution

We can break this problem down into parts and apply the theorem for convergent series to combine each part together.

### Step (1): Find the Sum of the First Term

The first term in the problem is a geometric series that can be simplified:

align}"): \begin{align}\sum_{k=1}^{\infty} \frac{4}{5^{k-1}} = \frac{4}{1-1/5} = 5.\end{align}

### Step (2): Find the Sum of the Second Term

{\begin{aligned}\sum _{k=1}^{\infty }{\frac {2^{k}}{3^{k}}}&=\sum _{k=2}^{\infty }{\Big (}{\frac {2}{3}}{\Big )}^{k-1}&&(2.1)\\&=\sum _{k=1}^{\infty }{\Big (}{\frac {2}{3}}{\Big )}^{k-1}-1&&(2.2)\\&={\frac {a}{1-r}}-1&&(2.3)\\&={\frac {1}{1-2/3}}-1&&(2.4)\\&=3-1\\&=2.\\\end{aligned}} ### Step (3): Combining Results

Combining results from Steps (1) and (2) yields

{\begin{aligned}\sum _{k=1}^{\infty }{\Big (}{\frac {4}{5^{k-1}}}-{\frac {2^{k}}{3^{k}}}{\Big )}=5-2=3.\end{aligned}} ## Discussion of Each Step

### Step (1)

The infinite series

align}"): \begin{align} \sum_{k=1}^{\infty} \frac{4}{5^{k-1}} = \sum_{k=1}^{\infty} 4\Big(\frac{1}{5}\Big)^{k-1} \end{align}

is a geometric series with common ratio $r=1/5$ and first term $a=4$ . Therefore, we can apply our formula

{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}={\frac {a}{1-r}}\end{aligned}} for computing the sum of a geometric series.

### Step (2.1)

The infinite series is geometric, and so we can find its sum by working it into the the form

{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}={\frac {a}{1-r}}\end{aligned}} to apply our summation formula. One way of starting this process is to making the exponent equal to $k-1$ , which we carried out by shifting the summation index up by one (other approaches may be used, see discussion below in the Possible Challenges section).

### Step (2.2)

The index of summation can be shifted down to 1, requiring that we subtract 1. This process was described in the lecture on sigma notation.

### Steps (2.3) and (2.4)

Now that our geometric series in the needed form to apply our summation formula, we may equate our series to

{\begin{aligned}{\frac {a}{1-r}}-1&&(2.3)\end{aligned}} and substitute our values of $r$ and $a$ to obtain Step (2.4).

The rest of Step (2) is algebraic manipulation (fractions).

### Step (3)

Because we have found two convergent infinite series, we can invoke the fourth property of convergent series (the sum of two convergent series is a convergent series) to compute the sum of the given problem:

{\begin{aligned}\sum _{k=1}^{\infty }{\Big (}{\frac {4}{5^{k-1}}}-{\frac {2^{k}}{3^{k}}}{\Big )}=5-2=3.\end{aligned}} For demonstration purposes, more steps were shown than what students may find that are needed to solve problems during assessments.

## Possible Challenge Areas

### Converting the Series Into the Needed Form

Students completing this problem may have noticed that our geometric series formula

{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}={\frac {a}{1-r}}\end{aligned}} can only be applied when the summation starts at 1 and the exponent is $k-1$ . In simplifying the second term, we needed to convert our problem into this form, which could be solved in another way. Another way of making the conversion is as follows:

{\begin{aligned}\sum _{k=1}^{\infty }{\frac {2^{k}}{3^{k}}}&=\sum _{k=1}^{\infty }{\frac {2^{k}}{3^{k}}}+(1-1)\\&={\Big [}\sum _{k=0}^{\infty }{\Big (}{\frac {2}{3}}{\Big )}^{k}{\Big ]}-1,\ \mathrm {absorb\ the\ positive\ 1\ into\ sum} \\&={\Big [}\sum _{k=1}^{\infty }{\Big (}{\frac {2}{3}}{\Big )}^{k-1}{\Big ]}-1,\ \mathrm {shift\ summation\ index\ up\ by\ 1} \\&={\frac {1}{1-2/3}}-1,\ \mathrm {using\ the\ summation\ formula\ for\ a\ convergent\ geometric\ series} \\&=2.\\\end{aligned}} Yet even more methods of finding the sum of this series could be used. While we have applied the formula

{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}={\frac {a}{1-r}}\end{aligned}} some calculus textbooks introduced the equivalent formula

{\begin{aligned}\sum _{k=0}^{\infty }ar^{k}={\frac {a}{1-r}}\end{aligned}} which leads to other similar approaches to finding the sum of the infinite series.