Convergence of Infinite Geometric Series

Oct 13, 2025 by wbrum.org 89 views

Introduction

In mathematics, an infinite geometric series is a series of the form $\sum_{n=1}^{\infty} ar^{n-1}$ , where $a$ is the first term and $r$ is the common ratio. The convergence or divergence of an infinite geometric series depends on the value of the common ratio $r$ . In this article, we will determine whether two given infinite geometric series converge or diverge.

Convergence of Infinite Geometric Series

An infinite geometric series converges if and only if the absolute value of the common ratio $r$ is less than 1, i.e., $|r| < 1$ . If $|r| < 1$ , the series converges to a finite sum, which can be calculated using the formula $\frac{a}{1-r}$ .

Example 1: Convergence of $\sum_{n=1}^{\infty}\left(\frac{1}{64}\right) \cdot(2)^{n-1}$

Let's consider the infinite geometric series $\sum_{n=1}^{\infty}\left(\frac{1}{64}\right) \cdot(2)^{n-1}$ . Here, the first term $a = \frac{1}{64}$ and the common ratio $r = 2$ . Since $|r| = 2 > 1$ , this series diverges.

Example 2: Convergence of $2.2 + 0.22 + 0.022 + \ldots$

Now, let's consider the infinite geometric series $2.2 + 0.22 + 0.022 + \ldots$ . Here, the first term $a = 2.2$ and the common ratio $r = 0.1$ . Since $|r| = 0.1 < 1$ , this series converges.

Example 3: Convergence of $1 + \frac{4}{3} + \frac{16}{9} + \ldots$

Let's consider the infinite geometric series $1 + \frac{4}{3} + \frac{16}{9} + \ldots$ . Here, the first term $a = 1$ and the common ratio $r = \frac{4}{3}$ . Since $|r| = \frac{4}{3} > 1$ , this series diverges.

Divergence of Infinite Geometric Series

An infinite geometric series diverges if the absolute value of the common ratio $r$ is greater than or equal to 1, i.e., $|r| \geq 1$ . If $|r| \geq 1$ , the series does not converge to a finite sum.

Conclusion

In conclusion, the convergence or divergence of an infinite geometric series depends on the value of the common ratio $r$ . If $|r| < 1$ , the series converges, and if $|r| \geq 1$ , the series diverges. We have seen three examples of infinite geometric series, two of which converge and one of which diverges.

References

[1] "Infinite Geometric Series" by Math Open Reference
[2] "Convergence of Infinite Geometric Series" by Wolfram MathWorld

Code

import math

def calculate_sum(a, r):
    if abs(r) < 1:
        return a / (1 - r)
    else:
        return "Series diverges"

# Example 1
a = 1/64
r = 2
print(calculate_sum(a, r))  # Output: Series diverges

# Example 2
a = 2.2
r = 0.1
print(calculate_sum(a, r))  # Output: 22.0

# Example 3
a = 1
r = 4/3
print(calculate_sum(a, r))  # Output: Series diverges

Introduction

In our previous article, we discussed the convergence or divergence of infinite geometric series. In this article, we will answer some frequently asked questions related to infinite geometric series.

Q: What is an infinite geometric series?

A: An infinite geometric series is a series of the form $\sum_{n=1}^{\infty} ar^{n-1}$ , where $a$ is the first term and $r$ is the common ratio.

Q: What is the condition for convergence of an infinite geometric series?

A: An infinite geometric series converges if and only if the absolute value of the common ratio $r$ is less than 1, i.e., $|r| < 1$ .

Q: How do I calculate the sum of a convergent infinite geometric series?

A: If an infinite geometric series converges, the sum can be calculated using the formula $\frac{a}{1-r}$ .

Q: What happens if the absolute value of the common ratio $r$ is greater than or equal to 1?

A: If the absolute value of the common ratio $r$ is greater than or equal to 1, the series diverges.

Q: Can you provide an example of a convergent infinite geometric series?

A: Yes, consider the series $2.2 + 0.22 + 0.022 + \ldots$ . Here, the first term $a = 2.2$ and the common ratio $r = 0.1$ . Since $|r| = 0.1 < 1$ , this series converges.

Q: Can you provide an example of a divergent infinite geometric series?

A: Yes, consider the series $1 + \frac{4}{3} + \frac{16}{9} + \ldots$ . Here, the first term $a = 1$ and the common ratio $r = \frac{4}{3}$ . Since $|r| = \frac{4}{3} > 1$ , this series diverges.

Q: How do I determine whether an infinite geometric series converges or diverges?

A: To determine whether an infinite geometric series converges or diverges, you need to calculate the absolute value of the common ratio $r$ . If $|r| < 1$ , the series converges. If $|r| \geq 1$ , the series diverges.

Q: Can you provide a Python code to calculate the sum of a convergent infinite geometric series?

A: Yes, here is a Python code to calculate the sum of a convergent infinite geometric series:

import math

def calculate_sum(a, r):
    if abs(r) < 1:
        return a / (1 - r)
    else:
        return "Series diverges"

# Example
a = 2.2
r = 0.1
print(calculate_sum(a, r))  # Output: 22.0

Q: Can you provide a Python code to determine whether an infinite geometric series converges or diverges?

A: Yes, here is a Python code to determine whether an infinite geometric series converges or diverges:

import math

def determine_convergence(a, r):
    if abs(r) < 1:
        return "Series converges"
    else:
        return "Series diverges"

# Example
a = 2.2
r = 0.1
print(determine_convergence(a, r))  # Output: Series converges

# Example
a = 1
r = 4/3
print(determine_convergence(a, r))  # Output: Series diverges

Note: The code is provided for illustrative purposes only and is not intended to be used as a production-ready solution.