infinite geometric sequence Archives

Geometric Sequences and Series

June 9, 2010 GB Calculus and Analysis, College Mathematics, High School Algebra, High School Mathematics

Introduction

We have discussed about arithmetic sequences, its characteristics and its connection to linear functions. In this post, we will discuss another type of sequence.

The sequence of numbers 2, 6, 18, 54, 162, … is an example of an geometric sequence. The first term 2 is multiplied by 3 to get the second term, the second term is multiplied by 3 to get the third term, the third term is multiplied by 3 to get the fourth term, and so on. The same number that we multiplied to each term is called the common ratio. Expressing the sequence above in terms of the first term and the common ratio, we have 2, 2(3), 2(3²), 2(3³), …. Hence, a geometric sequence, also known as a geometric progression, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.

The Sierpinski triangle below is an example of a geometric representation of a geometric sequence. The number of blue triangles, the number of white triangles, their areas, and their side lengths form different geometric sequences. It is left to the reader, as an exercise, to find the rules of these geometric sequences.

Figure 1 - The Seriepinski Triangles.

To generalize, if a₁ is its first term and the common ratio is r, then the general form of a geometric sequence is a₁, a₁r, a₁r², a₁r³,…, and the n^th term of the sequence is a₁r^n-1.

A geometric series, on the other hand, is the sum of the terms of a geometric sequence. Given a geometric sequence with terms a₁r, a₁r², a₁r³,…, the sum S_nof the geometric sequence with n terms is the geometric series a₁ + a₁r + a₁r², a₁r³ + … + ar^n-1. Multiplying S_n by -r and adding it to S_n, we have

Hence, the sum of a geometric series with n terms, and $r \neq 1 = \displaystyle\frac{a_1(1-r^n)}{1-r}$ .

Sum of Infinite Geometric Series and a Little Bit of Calculus

Note: This portion is for those who have already taken elementary calculus.

The infinite geometric series $\{a_n\}$ is the the symbol $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots$ . From above, the sum of a finite geometric series with $n$ terms is $\displaystyle \sum_{k=1}^n \frac{a_1(1-r^n)}{1-r}$ . Hence, to get the sum of the infinite geometric series, we need to get the sum of $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r}$ . However, $\displaystyle \sum_{k=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n\to \infty} \frac{a_1(1-r^n)}{1-r}$ .

Also, that if $|r| < 1$ , $r^n$ approaches $0$ (try $(\frac{2}{3})^n$ or any other proper fraction and increase the value of $n$ ), thus, $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r} = \frac{a_1}{1-r}$ . Therefore, sum of the infinite series $\displaystyle a_1 + a_2r + a_2r^2 + \cdots = \frac{a_1}{1-r}$ .

One very common infinite series is $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ , or the sum of the areas of the partitions of the square with side length 1 unit shown below. Using the formula above,

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{a_1}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$ .

Figure 2 - A representation of an infinite geometric series.

This is evident in the diagram because the sum of all the partitions is equal to the area of a square. We say that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n}$ converges to 1.

8 comments common ratio, geometric progressions, geometric sequence, infinite geometric sequence, sierpinski triangle, sum of infinite geometric series

Is 0.999… really equal to 1?

June 4, 2010 GB Calculus and Analysis, High School Calculus

Introduction

Yes it is. 0.999… is equal to 1.

Before we begin our discussion, let me make a remark that the symbol “…” in the decimal 0.999… means that the there are infinitely many 9’s, or putting it in plain language, the decimal number has no end.

A Friendly Chat About Whether 0.999… = 1 | BetterExplained via kwout

For non-math persons, you will probably disagree with the equality, but there are many elementary proofs that could show it, some of which, I have shown below. A proof is a series of valid, logical and relevant arguments (see Introduction to Mathematical Proofs for details), that shows the truth or falsity of a statement.

Proof 1

$\frac{1}{3} = 0.333 \cdots$

$\frac{2}{3} = 0.666 \cdots$

$\frac{1}{3} + \frac{2}{3} = 0.333 \cdots + 0.666 \cdots$

$\frac{3}{3} =0.999 \cdots$

But $\frac{3}{3} = 1$ , therefore $1 =0.999 \cdots$

Proof 2

$\frac{1}{9} = 0.111 \cdots$
Multiplying both sides by 9 we have

$1 = 0.999 \cdots$

Proof 3

Let $x = 0.999 \cdots$

$10x = 9.999 \cdots$

$10x - x = 9.999 \cdots - 0.9999 \cdots$

$9x = 9$

$x = 1$

Hence, $0.999 \cdots = 1$

Still in doubt?

Many will probably be reluctant in accepting the equality $1 = 0.999 \cdots$ because the representation is a bit counterintuitive. The said equality requires the notion of the real number system, a good grasp of the concept of limits, and knowledge on infinitesimals or calculus in general. If, for instance,you have already taken sequences (in calculus), you may think of the $0.999 \cdots$ as a sequence of real numbers $(0.9, 0.99, 0.999,\cdots)$ . Note that the sequence gets closer and closer to 1, and therefore, its limit is 1.

Infinite Geometric Sequence

My final attempt to convince you that $0.999 \cdots$ is indeed equal to $1$ is by the infinite geometric sequence. For the sake of brevity, in the remaining part of this article, we will simply use the term “infinite sequence” to refer to an infinite geometric sequence. We will use the concept of the sum of an infinite sequence, which is known as an infinite series, to show that $0.999 \cdots = 1$ .

One example of an infinite series is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ . If you add its infinite number of terms, the answer is equal to 1. Again, this is counterintuitive.

How can addition of numbers with infinite number of terms have an exact (or a finite) answer?

There is a formula to get the sum of an infinite geometric sequence, but before we discuss the formula, let me give the geometric interpretation of the sum above. The sum $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ can be represented geometrically using a 1 unit by 1 unit square as shown below. If we divide the square into two, then we will have two rectangles, each of which has area $\frac{1}{2}$ square units. Dividing the other half into two, then we have three rectangles with areas $\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{4}$ square units. Dividing the one of the smaller rectangle into two, then we have four rectangles with areas $\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{8}$ , $\frac{1}{8}$ . Again, dividing one of the smallest rectangle into two, we have five rectangles with areas $\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{8}$ , $\frac{1}{16}$ , and $\frac{1}{16}$ Since this process can go on forever, the sum of all the areas of all the rectangles will equal to 1, which is the area of the original square.

Now that we have seen that an infinite series can have a finite sum, we will now show that $0.999 \cdots$ can be expressed as a finite sum by expressing it as an infinite series. The number $0.999 \cdots$ can be expressed as an infinite series $0.9 + 0.09 + 0.009 + \cdots$ . Converting it in fractional form, we have $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$ .

We have learned that the sum of the infinite series with first term $\displaystyle a_1$ and ratio $r$ is described by $\displaystyle\frac{a_1}{1-r}$ . Applying the formula to our series above, we have

$\displaystyle\frac{\frac{9}{10}}{1-\frac{1}{10}} = 1$

Therefore, the sum our infinite series is 1.

Implication

This implication of the equality $0.999 \cdots =1$ means that any rational number that is a non-repeating decimal can be expressed as a repeating decimal. Since $0.999 \cdots =1$ , it follows that $0.0999 \cdots =0.1, 0.00999 \cdots=0.01$ and so on. Hence, any decimal number maybe expressed as number + 0.00…01. For example, the decimal $4.7$ , can be expressed as $4.6 + 0.1 = 4.6 + 0.0999 \cdots = 4.6999 \cdots$ . The number $0.874$ can also be expressed as $0.873 + 0.001 = 0.873 + 0.000999 \cdots = 0.873999 \cdots$

Conclusion

Any of the four proofs above is actually sufficient to show that $0.999 \cdots = 1$ . Although this concept is quite hard to accept, we should remember that in mathematics, as long as the steps of operations or reasoning performed are valid and logical, the conclusion will be unquestionably valid.

There are many counterintuitive concepts in mathematics and the equality $0.999 \cdots = 1$ is only one of the many. In my post, Counting the Uncountable: A Glimpse at the Infinite, we have also encountered one: that the number of integers (negative, 0, positive) is equal to the number of counting numbers (positive integers) and we have shown it by one-to-one pairing. We have also shown that the number of counting numbers is the same as the number of rational numbers. Thus, we have shown that a subset can have the same element as the “supposed” bigger set. I guess that is what makes mathematics unique; intuitively, some concepts do not make sense, but by valid and logical reasoning, they perfectly do.

Notes:

You can find discussions about 0.999… = 1 here and here.
There is another good post about it here and here.

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