Problem 27 - Algebra 2

Question

Find the sum of each infinite geometric series, if it exists. $$\sum_{n=1}^{\infty} \frac{1}{100}\left(\frac{101}{99}\right)^{n-1}$$

Step-by-Step Solution

Answer

The sum does not exist because the series diverges.

Step 1: Identify the first term and common ratio

The given series is of the form \( \sum_{n=1}^{\infty} a \cdot r^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio. For the series \( \sum_{n=1}^{\infty} \frac{1}{100}\left(\frac{101}{99}\right)^{n-1} \), we identify:- First term \( a = \frac{1}{100} \)- Common ratio \( r = \frac{101}{99} \) as it is used in the exponential term.

Step 2: Determine if the sum exists

An infinite geometric series converges to a sum if the common ratio \( r \) satisfies \( |r| < 1 \). In this case, \( r = \frac{101}{99} \), which is greater than 1, so \( |r| > 1 \). Therefore, this series does not converge.

Step 3: Conclude that the sum does not exist

Since the common ratio \( r \) is greater than 1, the series does not have a finite sum. Infinite geometric series only converge when \( |r| < 1 \). Hence, the sum of this series does not exist.

Key Concepts

Common Ratio

In an infinite geometric series, the common ratio is a crucial component that helps determine the behavior of the series. The common ratio, denoted as \( r \), is the factor by which we multiply each term to get the next term in the series. For example:

• If the first term is \( a \), the next term is \( a \times r \), the third term is \( a \times r^2 \), and so on.
• In the series \( \sum_{n=1}^{\infty} \frac{1}{100}\left(\frac{101}{99}\right)^{n-1} \), the common ratio is \( r = \frac{101}{99} \).

Understanding the common ratio is essential because it directly impacts whether the series converges or diverges. Specifically, if \( |r| < 1 \), the terms of the series get progressively smaller, allowing the series to converge to a specific sum. However, if \( |r| \geq 1 \), the terms either stay the same or increase, leading to divergence, meaning the series does not settle on a finite sum.

First Term

In any geometric series, identifying the first term is the starting point from which the entire series is built. The first term, often represented by \( a \), is the initial value of the sequence. It plays a foundational role in defining each subsequent term by being multiplied by powers of the common ratio, \( r \). For the given series, \( \sum_{n=1}^{\infty} \frac{1}{100}\left(\frac{101}{99}\right)^{n-1} \):

• The first term \( a = \frac{1}{100} \).

This first term is crucial because it sets the scale for the entire series. From here, each term is derived by applying the common ratio. Despite the structure of the series depending on both the first term and the common ratio, it is often the common ratio that determines whether the series converges.

Series Convergence

The concept of convergence in an infinite geometric series is vital to understanding whether the series adds up to a finite sum. Convergence is determined by the magnitude of the common ratio \( r \).

• The series converges if \( |r| < 1 \), meaning the absolute value of the common ratio is less than one. This causes the terms to diminish in size over time, allowing the sum to approach a specific value.
• If \( |r| \geq 1 \), the series diverges. This means the terms do not get smaller and thus can't settle on a finite value, instead, they increase or remain constant.

For the given series, \( \sum_{n=1}^{\infty} \frac{1}{100}\left(\frac{101}{99}\right)^{n-1} \), the common ratio \( r \) is \( \frac{101}{99} \), which is greater than 1. Hence, \( |r| > 1 \), and the series does not converge. Convergence criteria are essential because only converging series have a sum, making this property a central focus for determining possible solutions to geometric series problems.