Geometric series are a fundamental concept in mathematics, frequently encountered in various fields, including calculus, finance, and physics. A such series is a sequence of terms where each term is obtained by multiplying the previous term by a constant ratio.
In this article, we will delve into the properties of geometric series, discuss their convergence behavior, and explore methods for finding their sums.
Definition and Form
A geometric series is defined as the sum of the terms of a geometric sequence, where each term is obtained by multiplying the preceding term by a constant ratio. A geometric sequence takes the form a, ar, ar², ar³, …, where $a$ is the first term and $r$ is the common ratio. The sum of a such series can be expressed as $$ \sum_{n=0}^\infty a_n=a+ar+ar^2+ar^3+\cdots=\sum_{n=0}^\infty ar^n.$$
We define and discuss the convergence of the geometric series.
Convergence of Geometric Series
The convergence behavior of a geometric series is determined by the value of the common ratio $r$. The series converges when the absolute value of $r$ is less than 1, and it diverges otherwise.
Convergent Geometric Series
When |r| < 1, the geometric series converges, and its sum can be calculated using the formula S = a / (1 – r), where $S$ represents the sum of the series. This formula can be derived using the concept of limits and algebraic manipulation. It is important to note that the formula is valid only when |r| < 1.
Divergent of the Series
When |r| ≥ 1, the geometric series diverges, meaning it does not have a finite sum. In this case, the series grows indefinitely as more terms are added.
Examples of Geometric Series
Example 1: Consider the series $\sum_{n=0}^\infty$ 2ⁿ. Here, the first term ‘a’ is 1, and the common ratio ‘r’ is 2. As |r| = 2, which is greater than 1, the series diverges. To find the sum of the series, we use the formula S = a / (1 – r). Plugging in the values, we get S = 1 / (1 – 2) = -1. Therefore, the series $\sum_{n=0}^\infty$ 2ⁿ diverges.
Example 2: Consider the series $\sum_{n=0}^\infty$ (1/4)ⁿ. In this case, the first term ‘a’ is 1, and the common ratio ‘r’ is 1/4. As |r| = 1/4, which is less than 1, the series converges. To find the sum of the series, we use the formula S = a / (1 – r). Plugging in the values, we get S = 1 / (1 – 1/4) = 4/3. Therefore, the series $\sum_{n=0}^\infty$ (1/4)ⁿ converges, and its sum is 4/3.
We will see below that the convergence of the above series depends on the position of the real number $r$.
Applications to the convergence of series
Exercise: Prove the following series is convergent $$ \sum_{n=0}^\infty \sin(e^{-n}).$$ Proof: We recall that the Euler number $e>1$. So that $0<e^{-1}<n$. This implies that the series $\sum_{n=0}^{+\infty} e^{-n}$ is convergent, due to the previous paragraph. Using the property $|\sin(x)|\le |x|$ for any $x\in\mathbb{R},$ we have $$ |\sin(e^{-n})|\le e^{-n},\quad n\ge 0.$$ By using the comparison test for series of positive terms, we deduce that the series $ \sum_{n=0}^\infty \sin(e^{-n})$ is absolutely convergent, then it is a convergent series.
You may also consult the convergence of series using the ratio test for more examples and exercises with detailed answers.
