Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the defining characteristic of a geometric series? Give an example.

Short Answer

Expert verified
Answer: The defining characteristic of a geometric series is that each term is found by multiplying the previous term by a constant known as the common ratio (r). For example, in the geometric series 1, 2, 4, 8, 16, ..., each term is the result of multiplying the previous term by the common ratio of 2.

Step by step solution

01

Defining Characteristic of a Geometric Series

A geometric series is a sequence of numbers where each term is found by multiplying the previous term by a constant. This constant is called the common ratio, denoted by 'r'. In other words, every term in the series is a result of multiplying the preceding term by r, we can represent this as: term_n = term_{n-1} * r.
02

Example of a Geometric Series

Let's construct a simple geometric series with the common ratio, r = 2, and the first term as a_1 = 1. The series will look like: 1, 2, 4, 8, 16, ... Here, we can see that each term of the series is the result of multiplying the previous term by 2 (the common ratio). For example, the second term (2) is the product of the first term (1) and the common ratio (2), similarly, the third term (4) is the product of the second term (2) and the common ratio (2), and so on. This demonstrates the defining characteristic of a geometric series.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Ratio
Understanding the concept of the common ratio is essential when studying geometric series. A common ratio is the constant factor between consecutive terms in a geometric sequence. To find the common ratio, one simply divides any term by the previous term. For example, in the series 1, 2, 4, 8, 16, dividing the second term (2) by the first term (1) yields the common ratio of 2.
This ratio is 'common' because it is the same for all adjacent pairs of terms in the series. The formula representing this relationship is:
\[ term_{n} = term_{1} \times r^{(n-1)} \]
where \( term_{n} \) is the nth term in the series, \( term_{1} \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
To illustrate, for the fourth term in the series above:
\[ term_{4} = 1 \times 2^{(4-1)} = 8 \]
The common ratio is a driving force that gives the geometric series its exponential nature, which sharply contrasts with the linear nature of arithmetic series.
Arithmetic Sequence
In contrast to the multiplicative nature of geometric series, an arithmetic sequence is formed by adding a constant difference to each term to get the next term in the sequence. This constant difference is known as the 'common difference,' denoted by 'd'.
Consider the arithmetic sequence 3, 7, 11, 15, ... . The common difference here is 4, since each term is obtained by adding 4 to the previous term. The nth term of an arithmetic sequence is given by the formula:
\[ term_{n} = term_{1} + (n-1) \times d \]
where \( term_{1} \) is the first term, and \( n \) is the position of the term in the sequence. Thus, an arithmetic sequence is characterized by a consistent additive pattern, leading to a series that grows linearly rather than exponentially. This distinction is crucial for understanding the different ways in which sequences can evolve and is a fundamental part of sequence and series theory.
Convergent Series
A convergent series is one where the sum of its terms approaches a specific limit as more terms are added. This concept is especially important for geometric series, as their convergence depends on the value of the common ratio.
A geometric series converges if the absolute value of the common ratio is less than 1. In mathematical terms, if \( |r| < 1 \), then the series
\[ S = term_{1} + term_{1} \times r + term_{1} \times r^2 + ... \]
will approach a finite sum as the number of terms approaches infinity. The sum to infinity for a convergent geometric series can be calculated using the formula:
\[ S = \frac{term_{1}}{1 - r} \]
Back to our previous examples: a geometric series with a common ratio of 2 will not converge because \( |r| = 2 \) is not less than 1. Any geometric series with a common ratio with an absolute value greater than 1 will diverge, meaning the sum of its terms will grow infinitely large. Thus, understanding convergence is critical when analyzing the long-term behavior of series and determining if they have a finite sum.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The fractal called the snowflake island (or Koch island ) is constructed as follows: Let \(I_{0}\) be an equilateral triangle with sides of length \(1 .\) The figure \(I_{1}\) is obtained by replacing the middle third of each side of \(I_{0}\) with a new outward equilateral triangle with sides of length \(1 / 3\) (see figure). The process is repeated where \(I_{n+1}\) is obtained by replacing the middle third of each side of \(I_{n}\) with a new outward equilateral triangle with sides of length \(1 / 3^{n+1}\). The limiting figure as \(n \rightarrow \infty\) is called the snowflake island. a. Let \(L_{n}\) be the perimeter of \(I_{n} .\) Show that \(\lim _{n \rightarrow \infty} L_{n}=\infty\) b. Let \(A_{n}\) be the area of \(I_{n} .\) Find \(\lim _{n \rightarrow \infty} A_{n} .\) It exists!

For a positive real number \(p,\) the tower of exponents \(p^{p^{p}}\) continues indefinitely and the expression is ambiguous. The tower could be built from the top as the limit of the sequence \(\left\\{p^{p},\left(p^{p}\right)^{p},\left(\left(p^{p}\right)^{p}\right)^{p}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{1}=p^{p} .\) The tower could also be built from the bottom as the limit of the sequence \(\left\\{p^{p}, p^{\left(p^{p}\right)}, p^{\left(p^{(i)}\right)}, \ldots .\right\\},\) in which case the sequence is defined recursively as \(a_{n+1}=p^{a_{n}}(\text { building from the bottom })\) where again \(a_{1}=p^{p}\). a. Estimate the value of the tower with \(p=0.5\) by building from the top. That is, use tables to estimate the limit of the sequence defined recursively by (1) with \(p=0.5 .\) Estimate the maximum value of \(p > 0\) for which the sequence has a limit. b. Estimate the value of the tower with \(p=1.2\) by building from the bottom. That is, use tables to estimate the limit of the sequence defined recursively by (2) with \(p=1.2 .\) Estimate the maximum value of \(p > 1\) for which the sequence has a limit.

The Greeks solved several calculus problems almost 2000 years before the discovery of calculus. One example is Archimedes' calculation of the area of the region \(R\) bounded by a segment of a parabola, which he did using the "method of exhaustion." As shown in the figure, the idea was to fill \(R\) with an infinite sequence of triangles. Archimedes began with an isosceles triangle inscribed in the parabola, with area \(A_{1}\), and proceeded in stages, with the number of new triangles doubling at each stage. He was able to show (the key to the solution) that at each stage, the area of a new triangle is \(\frac{1}{8}\) of the area of a triangle at the previous stage; for example, \(A_{2}=\frac{1}{8} A_{1},\) and so forth. Show, as Archimedes did, that the area of \(R\) is \(\frac{4}{3}\) times the area of \(A_{1}\).

Consider the geometric series \(f(r)=\sum_{k=0}^{\infty} r^{k},\) where \(|r|<1\) a. Fill in the following table that shows the value of the series \(f(r)\) for various values of \(r\) $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline r & -0.9 & -0.7 & -0.5 & -0.2 & 0 & 0.2 & 0.5 & 0.7 & 0.9 \\ \hline f(r) & & & & & & & & & \\ \hline \end{array}$$ b. Graph \(f,\) for \(|r|<1\) \text { c. Evaluate } \lim _{r \rightarrow 1^{-}} f(r) \text { and } \lim _{r \rightarrow-1^{+}} f(r)

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) with terms that are non increasing in magnitude, converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n} .\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k .\) It can be shown that for \(n \geq 1\) \(\left|S-\left(S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right)\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|\) a. Interpret this inequality and explain why it is a better approximation to \(S\) than \(S_{n}\) b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free