Problem 10 - Essential Calculus Early Transcendentals

Question

Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \frac{10^{n}}{(-9)^{n-1}}$$

Step-by-Step Solution

Answer

The series is divergent because the common ratio's absolute value is greater than 1.

Step 1: Identify the first term and the common ratio

The given series is \( \sum_{n=1}^{\infty} \frac{10^n}{(-9)^{n-1}} \). To find the first term \(a\) and the common ratio \(r\), we rewrite the general term: \(a_n = \frac{10^n}{(-9)^{n-1}} = 10 \cdot \left( \frac{10}{-9} \right)^{n-1}\). Thus, the first term \(a = 10\) and the common ratio \(r = \frac{10}{-9} \).

Step 2: Check for convergence

For an infinite geometric series \( \sum a r^{n-1} \) to be convergent, the absolute value of the common ratio must be less than 1, i.e., \( |r| < 1 \). Here, \( |r| = \left| \frac{10}{-9} \right| = \frac{10}{9} \), which is greater than 1. Therefore, the series is divergent.

Key Concepts

Convergence and Divergence

In the context of geometric series, the terms convergence and divergence are key. They help us understand whether the series sums to a finite value or not. For a geometric series, we look at the common ratio, denoted as \( r \). If the absolute value \( |r| \) is less than 1, the series converges, implying it adds up to a specific number. This occurs because the terms become smaller and smaller, approaching zero. If \( |r| \) is equal to or greater than 1, the series diverges, meaning the sums of the infinite number of terms grow without bound and do not settle to a single value. In our specific exercise, the series is found to be divergent because the common ratio \( |r| = \frac{10}{9} \), which is greater than 1. Thus, the series does not converge to a finite sum.

Infinite Series

An infinite series is essentially a sum of an infinite sequence of numbers. These series play a crucial role in various areas of mathematics, including calculus and analysis. An infinite series can be written as \( S = a_1 + a_2 + a_3 + \, ... \), where \( a_n \) represents the terms of the series.In the context of a geometric series, each subsequent term is found by multiplying the previous term by the common ratio \( r \). The series continues indefinitely. However, the question of whether these sums add up (converge) or not (diverge) depends heavily on the common ratio. The series explored in the exercise \( \sum_{n=1}^{\infty} \frac{10^n}{(-9)^{n-1}} \) is an example of an infinite series known for its potential to demonstrate convergence or divergence depending on the value of \( r \).

Common Ratio

The common ratio is a cornerstone concept for understanding geometric series. It is the constant factor between consecutive terms of the series. Given a general geometric series \( a_n = ar^{n-1} \), \( r \) is the common ratio. You find \( r \) by dividing any term by its preceding term. It's represented as \( \frac{a_{n+1}}{a_n} \).In the exercise, the common ratio \( r \) is \( \frac{10}{-9} \), calculated from the formula. The sign of the ratio can influence the behavior of the series significantly. Here, even though \( r \) is negative, its absolute value being greater than 1 leads to divergence. Recognizing the common ratio quickly helps in determining if a series will converge or diverge. Knowing this can save time in analyzing complex series.