# Geometric Series

TOPIC: GEOMETRIC SERIES

Sub-topics:

• Convergent Geometric Series
• Divergent Geometric Series
• Word Problems Involving Geometric Series

LEARNING OBJECTIVES:

• To define geometric series and related concepts
• To differentiate convergent and divergent series
• To solve problems involving geometric series such as investment and compound interest

MATH CONCEPTS

• GEOMETRIC SERIES – It is the sum of the terms of a geometric sequence.
• CONVERGENT GEOMETRIC SERIES – A series that converges has a finite limit, that is a number that is approached. If |r| < 1, then the series will converge.
• DIVERGENT GEOMETRIC SERIES – A series that diverges means either the partial sums have no limit or approach infinity. If |r| > 1 then the series diverges.
• FORMULA IN GETTING THE SUM OF GEOMETRIC SERIES
• $\sum_{i=1}^{\infty }a_{i} =(\frac{a}{1-r})$ if $\left | r \right | \geq 1$ (CONVERGENT GEOMETRIC SERIES)
• $\sum_{i=1}^{n }a_{i} =a(\frac{1-r^{n}}{1-r})$ if $\left | r \right | \leq 1$ (DIVERGENT GEOMETRIC SERIES)

IMPORTANCE

• Understanding the concepts behind geometric series is a good foundation in learning about investment growth and loss.
• In business, the idea geometric sequence and series is used to represent and predict data.
• In higher math, series plays an important role in limits in Calculus.

DISCUSSION

GEOMETRIC SERIES

Geometric series is the sum of the terms in a geometric sequence. Remember that a geometric sequence has a common ratio. It is possible to take the sum of a finite number of terms of a geometric sequence. Also, when studying for calculus, you can take the sum of an infinite geometric sequence, but only in the specific condition that the common ratio r is between –1 and 1; that is, you have to have | r | < 1.

In general, if $a_{k+1}/a_{k}$ equal to constant r, the terms are of the form $a_{1}=a$ then the geometric sequence is given by: $a,\,ar,\,ar^{2},\,ar^{3},\,ar^{4}, ....,\,ar^{n}$

To get the sum, $S_{n} = \sum_{k=1}^{n}a_{k}=a+ar+ar^{2}+ar^{3}+ar^{4}+....+ar^{n-1}$   (equation 1)

If we multiply the given equation by r both sides, $(r)S_{n} = (a+ar+ar^{2}+ar^{3}+ar^{4}+....+ar^{n})(r)=rS_{n}=ar+ar^{2}+ar^{3}+ar^{4}+....+ar^{n}$    (equation 2)

By subtracting equation 1 from equation 2, $(rS_{n}=ar+ar^{2}+ar^{3}+ar^{4}+....+ar^{n})-(a+ar+ar^{2}+ar^{3}+ar^{4}+....+ar^{n-1})$ $(1-r)S_{n}=a-ar^{n}$

Dividing both sides by (1 – r), $S_{n}=\frac{(a-r^{n})}{(a-r)}$ or $S_{n}=a\frac{(1-r^{n})}{(1-r)}$ if $\left | r \right | \geq 1$

For series with infinite sum, $\sum_{i=1}^{\infty }a_{i} =(\frac{a}{1-r})$, if $\left | r \right | \leq 1$

Depending on the nature of the common ratio, r, a geometric series can converge or diverge. A series that converges has a finite limit, that is a number that is approached. If |r| < 1, then the series will converge. A series that diverges means either the partial sums have no limit or approach infinity. If $\left | r \right | \geq 1$ then the series diverges.

ILLUSTRATIVE EXAMPLES

Identify whether the following series converge or diverge then find the value of the following geometric series.

1. Find the sum of the first nine terms of the geometric sequence whose first term is -25 and the common ratio is 1.5.
2. Find $S_{15}$ of the sequence 1, 1/5, 1/25, 1/125, …
3. Find the sum of the first 12 terms of the following geometric progression: 2, 4, 8, 16, 32
4. $\sum_{i=1}^{25}3(-2)^{i}$

SOLUTIONS:

1) In finding the sum, identify first if the series converges or diverges. Since r = 1.5, the series diverges. Use the formula: $S_{n}=a\frac{(1-r^{n})}{(1-r)}$, where a = the first term, r = common ratio, and n = number of included terms.

a = -25        r = 1.5      n = 9 $S_{9}=(-25)\frac{[1-(1.5)^{9}]}{(1-1.5)}=(-25)\frac{[1-(1.5)^{9}]}{(-0.5)}=-1,872.168$

2) Since r = 1/5, the series converges. Use the formula: $\sum_{i=1}^{\infty }a_{i} =(\frac{a}{1-r})$, where a = the first term and r = common ratio.

a = 1            r = 15 $a_{15} =(\frac{1}{1-\frac{1}{5}})=\frac{1}{(\frac{4}{5})}=\frac{5}{4}$

3) Since r = 2, the series diverges. Use the formula: $S_{n}=a\frac{(1-r^{n})}{(1-r)}$, where a = the first term, r = common ratio, and n = number of included terms.

a = 2        r = 2        n = 12 $S_{12}=(2)\frac{[1-(2)^{12}]}{(1-2)}=(2)\frac{[1-(2)^{12}]}{(-1)}=8,190$

4) Since r = -2, the series diverges. Use the formula: $S_{n}=a\frac{(1-r^{n})}{(1-r)}$, where a = the first term, r = common ratio, and n = number of included terms. $a=3(-2^{1})=-6$        r = -2        n = 25 $S_{25}=(-6)\frac{[1-(-2)^{25}]}{(1-(-2))}=(-6)\frac{[1-(-2)^{25}]}{3}= -67,108,866$

DISCUSSION

REAL-LIFE APPLICATIONS OF GEOMETRIC SERIES

Do you know that archeologists in the movies, such as Indiana Jones, can estimate the age of different artifacts? Do not you know that the age of artifacts in real life can be established by the amount of the radioactive isotope of Carbon 14 in the artifact? Carbon 14 has a very long half-lifetime which means that each half-lifetime of 5730 years or so, the amount of the isotope is reduced by half. Hence, these consecutive amounts of Carbon 14 are the terms of a decreasing geometric progression with a common ratio of ½. Geometric series is also used to predict a final amount of money that is invested in a certain period of time. Also, using geometric series, we can determine the increase and decrease of population of a particular city. Geometric series has plenty of applications in real life.

ILLUSTRATIVE EXAMPLES

Solve the following word problems.

1. In 2000, the number of births of a certain town was 800. A pattern was observed that the number is increasing by 1.2 % each year.  What is the total number of births over a 20 year period to a population that grows at a fixed percentage each year?
2. You want to invest your money in a bank that is offering a 2.3% compound interest each year. If you invested \$5,000 in this bank for six years, calculate the total sum of the investment growth.

SOLUTIONS:

1) Let’s list down the first 3 terms of the sequence with the first term as 800 and the rate of increase is 1.2%.

800 x 1.012 = 809.6, 810 x 1.012 = 819.32, 819.32 x 1.012 = 829.15 $S_{20}=(800)\frac{[1-(1.012)^{20}]}{(1-(1.012))}=(800)\frac{[1-(1.012)^{20}]}{(-0.012)}=17,962.29$  or 17,926 number of births

2) The first term is 5000 and the rate of increase is 2.3%. $S_{6}=(5000)\frac{[1-(1.023)^{6}]}{(1-(1.023))}=(5000)\frac{[1-(1.023)^{6}]}{(-0.023)}=31,778.82$

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